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This question already has answers here:
How do you use a variable in a regular expression?
(27 answers)
Closed 2 years ago.
So for example:
function(input){
var testVar = input;
string = ...
string.replace(/ReGeX + testVar + ReGeX/, "replacement")
}
But this is of course not working :)
Is there any way to do this?
const regex = new RegExp(`ReGeX${testVar}ReGeX`);
...
string.replace(regex, "replacement");
Update
Per some of the comments, it's important to note that you may want to escape the variable if there is potential for malicious content (e.g. the variable comes from user input)
ES6 Update
In 2019, this would usually be written using a template string, and the above code has been updated. The original answer was:
var regex = new RegExp("ReGeX" + testVar + "ReGeX");
...
string.replace(regex, "replacement");
You can use the RegExp object:
var regexstring = "whatever";
var regexp = new RegExp(regexstring, "gi");
var str = "whateverTest";
var str2 = str.replace(regexp, "other");
document.write(str2);
Then you can construct regexstring in any way you want.
You can read more about it here.
To build a regular expression from a variable in JavaScript, you'll need to use the RegExp constructor with a string parameter.
function reg(input) {
var flags;
//could be any combination of 'g', 'i', and 'm'
flags = 'g';
return new RegExp('ReGeX' + input + 'ReGeX', flags);
}
of course, this is a very naive example. It assumes that input is has been properly escaped for a regular expression. If you're dealing with user-input, or simply want to make it more convenient to match special characters, you'll need to escape special characters:
function regexEscape(str) {
return str.replace(/[-\/\\^$*+?.()|[\]{}]/g, '\\$&')
}
function reg(input) {
var flags;
//could be any combination of 'g', 'i', and 'm'
flags = 'g';
input = regexEscape(input);
return new RegExp('ReGeX' + input + 'ReGeX', flags);
}
You can create regular expressions in JS in one of two ways:
Using regular expression literal - /ab{2}/g
Using the regular expression constructor - new RegExp("ab{2}", "g") .
Regular expression literals are constant, and can not be used with variables. This could be achieved using the constructor. The stracture of the RegEx constructor is
new RegExp(regularExpressionString, modifiersString)
You can embed variables as part of the regularExpressionString. For example,
var pattern="cd"
var repeats=3
new RegExp(`${pattern}{${repeats}}`, "g")
This will match any appearance of the pattern cdcdcd.
if you're using es6 template literals are an option...
string.replace(new RegExp(`ReGeX${testVar}ReGeX`), "replacement")
You can always give regular expression as string, i.e. "ReGeX" + testVar + "ReGeX". You'll possibly have to escape some characters inside your string (e.g., double quote), but for most cases it's equivalent.
You can also use RegExp constructor to pass flags in (see the docs).
It's only necessary to prepare the string variable first and then convert it to the RegEx.
for example:
You want to add minLength and MaxLength with the variable to RegEx:
function getRegEx() {
const minLength = "5"; // for exapmle: min is 5
const maxLength = "12"; // for exapmle: man is 12
var regEx = "^.{" + minLength + ","+ maxLength +"}$"; // first we make a String variable of our RegEx
regEx = new RegExp(regEx, "g"); // now we convert it to RegEx
return regEx; // In the end, we return the RegEx
}
now if you change value of MaxLength or MinLength, It will change in all RegExs.
Hope to be useful. Also sorry about my English.
Here's an pretty useless function that return values wrapped by specific characters. :)
jsfiddle: https://jsfiddle.net/squadjot/43agwo6x/
function getValsWrappedIn(str,c1,c2){
var rg = new RegExp("(?<=\\"+c1+")(.*?)(?=\\"+c2+")","g");
return str.match(rg);
}
var exampleStr = "Something (5) or some time (19) or maybe a (thingy)";
var results = getValsWrappedIn(exampleStr,"(",")")
// Will return array ["5","19","thingy"]
console.log(results)
accepted answer doesn't work for me and doesn't follow MDN examples
see the 'Description' section in above link
I'd go with the following it's working for me:
let stringThatIsGoingToChange = 'findMe';
let flagsYouWant = 'gi' //simple string with flags
let dynamicRegExp = new RegExp(`${stringThatIsGoingToChange}`, flagsYouWant)
// that makes dynamicRegExp = /findMe/gi
Given a string:
s = "Test abc test test abc test test test abc test test abc";
This seems to only remove the first occurrence of abc in the string above:
s = s.replace('abc', '');
How do I replace all occurrences of it?
As of August 2020: Modern browsers have support for the String.replaceAll() method defined by the ECMAScript 2021 language specification.
For older/legacy browsers:
function escapeRegExp(string) {
return string.replace(/[.*+?^${}()|[\]\\]/g, '\\$&'); // $& means the whole matched string
}
function replaceAll(str, find, replace) {
return str.replace(new RegExp(escapeRegExp(find), 'g'), replace);
}
Here is how this answer evolved:
str = str.replace(/abc/g, '');
In response to comment "what's if 'abc' is passed as a variable?":
var find = 'abc';
var re = new RegExp(find, 'g');
str = str.replace(re, '');
In response to Click Upvote's comment, you could simplify it even more:
function replaceAll(str, find, replace) {
return str.replace(new RegExp(find, 'g'), replace);
}
Note: Regular expressions contain special (meta) characters, and as such it is dangerous to blindly pass an argument in the find function above without pre-processing it to escape those characters. This is covered in the Mozilla Developer Network's JavaScript Guide on Regular Expressions, where they present the following utility function (which has changed at least twice since this answer was originally written, so make sure to check the MDN site for potential updates):
function escapeRegExp(string) {
return string.replace(/[.*+?^${}()|[\]\\]/g, '\\$&'); // $& means the whole matched string
}
So in order to make the replaceAll() function above safer, it could be modified to the following if you also include escapeRegExp:
function replaceAll(str, find, replace) {
return str.replace(new RegExp(escapeRegExp(find), 'g'), replace);
}
For the sake of completeness, I got to thinking about which method I should use to do this. There are basically two ways to do this as suggested by the other answers on this page.
Note: In general, extending the built-in prototypes in JavaScript is generally not recommended. I am providing as extensions on the String prototype simply for purposes of illustration, showing different implementations of a hypothetical standard method on the String built-in prototype.
Regular Expression Based Implementation
String.prototype.replaceAll = function(search, replacement) {
var target = this;
return target.replace(new RegExp(search, 'g'), replacement);
};
Split and Join (Functional) Implementation
String.prototype.replaceAll = function(search, replacement) {
var target = this;
return target.split(search).join(replacement);
};
Not knowing too much about how regular expressions work behind the scenes in terms of efficiency, I tended to lean toward the split and join implementation in the past without thinking about performance. When I did wonder which was more efficient, and by what margin, I used it as an excuse to find out.
On my Chrome Windows 8 machine, the regular expression based implementation is the fastest, with the split and join implementation being 53% slower. Meaning the regular expressions are twice as fast for the lorem ipsum input I used.
Check out this benchmark running these two implementations against each other.
As noted in the comment below by #ThomasLeduc and others, there could be an issue with the regular expression-based implementation if search contains certain characters which are reserved as special characters in regular expressions. The implementation assumes that the caller will escape the string beforehand or will only pass strings that are without the characters in the table in Regular Expressions (MDN).
MDN also provides an implementation to escape our strings. It would be nice if this was also standardized as RegExp.escape(str), but alas, it does not exist:
function escapeRegExp(str) {
return str.replace(/[.*+?^${}()|[\]\\]/g, "\\$&"); // $& means the whole matched string
}
We could call escapeRegExp within our String.prototype.replaceAll implementation, however, I'm not sure how much this will affect the performance (potentially even for strings for which the escape is not needed, like all alphanumeric strings).
In the latest versions of most popular browsers, you can use replaceAll
as shown here:
let result = "1 abc 2 abc 3".replaceAll("abc", "xyz");
// `result` is "1 xyz 2 xyz 3"
But check Can I use or another compatibility table first to make sure the browsers you're targeting have added support for it first.
For Node.js and compatibility with older/non-current browsers:
Note: Don't use the following solution in performance critical code.
As an alternative to regular expressions for a simple literal string, you could use
str = "Test abc test test abc test...".split("abc").join("");
The general pattern is
str.split(search).join(replacement)
This used to be faster in some cases than using replaceAll and a regular expression, but that doesn't seem to be the case anymore in modern browsers.
Benchmark: https://jsben.ch/TZYzj
Conclusion:
If you have a performance-critical use case (e.g., processing hundreds of strings), use the regular expression method. But for most typical use cases, this is well worth not having to worry about special characters.
Here's a string prototype function based on the accepted answer:
String.prototype.replaceAll = function(find, replace) {
var str = this;
return str.replace(new RegExp(find, 'g'), replace);
};
If your find contains special characters then you need to escape them:
String.prototype.replaceAll = function(find, replace) {
var str = this;
return str.replace(new RegExp(find.replace(/[-\/\\^$*+?.()|[\]{}]/g, '\\$&'), 'g'), replace);
};
Fiddle: http://jsfiddle.net/cdbzL/
Use word boundaries (\b)
'a cat is not a caterpillar'.replace(/\bcat\b/gi,'dog');
//"a dog is not a caterpillar"
This is a simple regex that avoids replacing parts of words in most cases. However, a dash - is still considered a word boundary. So conditionals can be used in this case to avoid replacing strings like cool-cat:
'a cat is not a cool-cat'.replace(/\bcat\b/gi,'dog');//wrong
//"a dog is not a cool-dog" -- nips
'a cat is not a cool-cat'.replace(/(?:\b([^-]))cat(?:\b([^-]))/gi,'$1dog$2');
//"a dog is not a cool-cat"
Basically, this question is the same as the question here:
Replace " ' " with " '' " in JavaScript
Regexp isn't the only way to replace multiple occurrences of a substring, far from it. Think flexible, think split!
var newText = "the cat looks like a cat".split('cat').join('dog');
Alternatively, to prevent replacing word parts—which the approved answer will do, too! You can get around this issue using regular expressions that are, I admit, somewhat more complex and as an upshot of that, a tad slower, too:
var regText = "the cat looks like a cat".replace(/(?:(^|[^a-z]))(([^a-z]*)(?=cat)cat)(?![a-z])/gi,"$1dog");
The output is the same as the accepted answer, however, using the /cat/g expression on this string:
var oops = 'the cat looks like a cat, not a caterpillar or coolcat'.replace(/cat/g,'dog');
//returns "the dog looks like a dog, not a dogerpillar or cooldog" ??
Oops indeed, this probably isn't what you want. What is, then? IMHO, a regex that only replaces 'cat' conditionally (i.e., not part of a word), like so:
var caterpillar = 'the cat looks like a cat, not a caterpillar or coolcat'.replace(/(?:(^|[^a-z]))(([^a-z]*)(?=cat)cat)(?![a-z])/gi,"$1dog");
//return "the dog looks like a dog, not a caterpillar or coolcat"
My guess is, this meets your needs. It's not foolproof, of course, but it should be enough to get you started. I'd recommend reading some more on these pages. This'll prove useful in perfecting this expression to meet your specific needs.
RegExp (regular expression) object
Regular-Expressions.info
Here is an example of .replace used with a callback function. In this case, it dramatically simplifies the expression and provides even more flexibility, like replacing with correct capitalisation or replacing both cat and cats in one go:
'Two cats are not 1 Cat! They\'re just cool-cats, you caterpillar'
.replace(/(^|.\b)(cat)(s?\b.|$)/gi,function(all,char1,cat,char2)
{
// Check 1st, capitalize if required
var replacement = (cat.charAt(0) === 'C' ? 'D' : 'd') + 'og';
if (char1 === ' ' && char2 === 's')
{ // Replace plurals, too
cat = replacement + 's';
}
else
{ // Do not replace if dashes are matched
cat = char1 === '-' || char2 === '-' ? cat : replacement;
}
return char1 + cat + char2;//return replacement string
});
//returns:
//Two dogs are not 1 Dog! They're just cool-cats, you caterpillar
These are the most common and readable methods.
var str = "Test abc test test abc test test test abc test test abc"
Method 1:
str = str.replace(/abc/g, "replaced text");
Method 2:
str = str.split("abc").join("replaced text");
Method 3:
str = str.replace(new RegExp("abc", "g"), "replaced text");
Method 4:
while(str.includes("abc")){
str = str.replace("abc", "replaced text");
}
Output:
console.log(str);
// Test replaced text test test replaced text test test test replaced text test test replaced text
Match against a global regular expression:
anotherString = someString.replace(/cat/g, 'dog');
For replacing a single time, use:
var res = str.replace('abc', "");
For replacing multiple times, use:
var res = str.replace(/abc/g, "");
str = str.replace(/abc/g, '');
Or try the replaceAll method, as recommended in this answer:
str = str.replaceAll('abc', '');
or:
var search = 'abc';
str = str.replaceAll(search, '');
EDIT: Clarification about replaceAll availability
The replaceAll method is added to String's prototype. This means it will be available for all string objects/literals.
Example:
var output = "test this".replaceAll('this', 'that'); // output is 'test that'.
output = output.replaceAll('that', 'this'); // output is 'test this'
Using RegExp in JavaScript could do the job for you. Just simply do something like below code, and don't forget the /g after which standout for global:
var str ="Test abc test test abc test test test abc test test abc";
str = str.replace(/abc/g, '');
If you think of reuse, create a function to do that for you, but it's not recommended as it's only one line function. But again, if you heavily use this, you can write something like this:
String.prototype.replaceAll = String.prototype.replaceAll || function(string, replaced) {
return this.replace(new RegExp(string, 'g'), replaced);
};
And simply use it in your code over and over like below:
var str ="Test abc test test abc test test test abc test test abc";
str = str.replaceAll('abc', '');
But as I mention earlier, it won't make a huge difference in terms of lines to be written or performance. Only caching the function may affect some faster performance on long strings and is also a good practice of DRY code if you want to reuse.
Say you want to replace all the 'abc' with 'x':
let some_str = 'abc def def lom abc abc def'.split('abc').join('x')
console.log(some_str) //x def def lom x x def
I was trying to think about something more simple than modifying the string prototype.
Use a regular expression:
str.replace(/abc/g, '');
Performance
Today 27.12.2019 I perform tests on macOS v10.13.6 (High Sierra) for the chosen solutions.
Conclusions
The str.replace(/abc/g, ''); (C) is a good cross-browser fast solution for all strings.
Solutions based on split-join (A,B) or replace (C,D) are fast
Solutions based on while (E,F,G,H) are slow - usually ~4 times slower for small strings and about ~3000 times (!) slower for long strings
The recurrence solutions (RA,RB) are slow and do not work for long strings
I also create my own solution. It looks like currently it is the shortest one which does the question job:
str.split`abc`.join``
str = "Test abc test test abc test test test abc test test abc";
str = str.split`abc`.join``
console.log(str);
Details
The tests were performed on Chrome 79.0, Safari 13.0.4 and Firefox 71.0 (64 bit). The tests RA and RB use recursion. Results
Short string - 55 characters
You can run tests on your machine HERE. Results for Chrome:
Long string: 275 000 characters
The recursive solutions RA and RB gives
RangeError: Maximum call stack size exceeded
For 1M characters they even break Chrome
I try to perform tests for 1M characters for other solutions, but E,F,G,H takes so much time that browser ask me to break script so I shrink test string to 275K characters. You can run tests on your machine HERE. Results for Chrome
Code used in tests
var t="Test abc test test abc test test test abc test test abc"; // .repeat(5000)
var log = (version,result) => console.log(`${version}: ${result}`);
function A(str) {
return str.split('abc').join('');
}
function B(str) {
return str.split`abc`.join``; // my proposition
}
function C(str) {
return str.replace(/abc/g, '');
}
function D(str) {
return str.replace(new RegExp("abc", "g"), '');
}
function E(str) {
while (str.indexOf('abc') !== -1) { str = str.replace('abc', ''); }
return str;
}
function F(str) {
while (str.indexOf('abc') !== -1) { str = str.replace(/abc/, ''); }
return str;
}
function G(str) {
while(str.includes("abc")) { str = str.replace('abc', ''); }
return str;
}
// src: https://stackoverflow.com/a/56989553/860099
function H(str)
{
let i = -1
let find = 'abc';
let newToken = '';
if (!str)
{
if ((str == null) && (find == null)) return newToken;
return str;
}
while ((
i = str.indexOf(
find, i >= 0 ? i + newToken.length : 0
)) !== -1
)
{
str = str.substring(0, i) +
newToken +
str.substring(i + find.length);
}
return str;
}
// src: https://stackoverflow.com/a/22870785/860099
function RA(string, prevstring) {
var omit = 'abc';
var place = '';
if (prevstring && string === prevstring)
return string;
prevstring = string.replace(omit, place);
return RA(prevstring, string)
}
// src: https://stackoverflow.com/a/26107132/860099
function RB(str) {
var find = 'abc';
var replace = '';
var i = str.indexOf(find);
if (i > -1){
str = str.replace(find, replace);
i = i + replace.length;
var st2 = str.substring(i);
if(st2.indexOf(find) > -1){
str = str.substring(0,i) + RB(st2, find, replace);
}
}
return str;
}
log('A ', A(t));
log('B ', B(t));
log('C ', C(t));
log('D ', D(t));
log('E ', E(t));
log('F ', F(t));
log('G ', G(t));
log('H ', H(t));
log('RA', RA(t)); // use reccurence
log('RB', RB(t)); // use reccurence
<p style="color:red">This snippet only presents codes used in tests. It not perform test itself!<p>
Replacing single quotes:
function JavaScriptEncode(text){
text = text.replace(/'/g,''')
// More encode here if required
return text;
}
Using
str = str.replace(new RegExp("abc", 'g'), "");
worked better for me than the previous answers. So new RegExp("abc", 'g') creates a regular expression what matches all occurrences ('g' flag) of the text ("abc"). The second part is what gets replaced to, in your case empty string ("").
str is the string, and we have to override it, as replace(...) just returns result, but not overrides. In some cases you might want to use that.
This is the fastest version that doesn't use regular expressions.
Revised jsperf
replaceAll = function(string, omit, place, prevstring) {
if (prevstring && string === prevstring)
return string;
prevstring = string.replace(omit, place);
return replaceAll(prevstring, omit, place, string)
}
It is almost twice as fast as the split and join method.
As pointed out in a comment here, this will not work if your omit variable contains place, as in: replaceAll("string", "s", "ss"), because it will always be able to replace another occurrence of the word.
There is another jsperf with variants on my recursive replace that go even faster (http://jsperf.com/replace-all-vs-split-join/12)!
Update July 27th 2017: It looks like RegExp now has the fastest performance in the recently released Chrome 59.
Loop it until number occurrences comes to 0, like this:
function replaceAll(find, replace, str) {
while (str.indexOf(find) > -1) {
str = str.replace(find, replace);
}
return str;
}
If what you want to find is already in a string, and you don't have a regex escaper handy, you can use join/split:
function replaceMulti(haystack, needle, replacement)
{
return haystack.split(needle).join(replacement);
}
someString = 'the cat looks like a cat';
console.log(replaceMulti(someString, 'cat', 'dog'));
function replaceAll(str, find, replace) {
var i = str.indexOf(find);
if (i > -1){
str = str.replace(find, replace);
i = i + replace.length;
var st2 = str.substring(i);
if(st2.indexOf(find) > -1){
str = str.substring(0,i) + replaceAll(st2, find, replace);
}
}
return str;
}
I like this method (it looks a little cleaner):
text = text.replace(new RegExp("cat","g"), "dog");
String.prototype.replaceAll - ECMAScript 2021
The new String.prototype.replaceAll() method returns a new string with all matches of a pattern replaced by a replacement. The pattern can be either a string or a RegExp, and the replacement can be either a string or a function to be called for each match.
const message = 'dog barks meow meow';
const messageFormatted = message.replaceAll('meow', 'woof')
console.log(messageFormatted);
Of course in 2021 the right answer is:
String.prototype.replaceAll()
console.log(
'Change this and this for me'.replaceAll('this','that') // Normal case
);
console.log(
'aaaaaa'.replaceAll('aa','a') // Challenged case
);
If you don't want to deal with replace() + RegExp.
But what if the browser is from before 2020?
In this case we need polyfill (forcing older browsers to support new features) (I think for a few years will be necessary).
I could not find a completely right method in answers. So I suggest this function that will be defined as a polyfill.
My suggested options for replaceAll polyfill:
replaceAll polyfill (with global-flag error) (more principled version)
if (!String.prototype.replaceAll) { // Check if the native function not exist
Object.defineProperty(String.prototype, 'replaceAll', { // Define replaceAll as a prototype for (Mother/Any) String
configurable: true, writable: true, enumerable: false, // Editable & non-enumerable property (As it should be)
value: function(search, replace) { // Set the function by closest input names (For good info in consoles)
return this.replace( // Using native String.prototype.replace()
Object.prototype.toString.call(search) === '[object RegExp]' // IsRegExp?
? search.global // Is the RegEx global?
? search // So pass it
: function(){throw new TypeError('replaceAll called with a non-global RegExp argument')}() // If not throw an error
: RegExp(String(search).replace(/[.^$*+?()[{|\\]/g, "\\$&"), "g"), // Replace all reserved characters with '\' then make a global 'g' RegExp
replace); // passing second argument
}
});
}
replaceAll polyfill (With handling global-flag missing by itself) (my first preference) - Why?
if (!String.prototype.replaceAll) { // Check if the native function not exist
Object.defineProperty(String.prototype, 'replaceAll', { // Define replaceAll as a prototype for (Mother/Any) String
configurable: true, writable: true, enumerable: false, // Editable & non-enumerable property (As it should be)
value: function(search, replace) { // Set the function by closest input names (For good info in consoles)
return this.replace( // Using native String.prototype.replace()
Object.prototype.toString.call(search) === '[object RegExp]' // IsRegExp?
? search.global // Is the RegEx global?
? search // So pass it
: RegExp(search.source, /\/([a-z]*)$/.exec(search.toString())[1] + 'g') // If not, make a global clone from the RegEx
: RegExp(String(search).replace(/[.^$*+?()[{|\\]/g, "\\$&"), "g"), // Replace all reserved characters with '\' then make a global 'g' RegExp
replace); // passing second argument
}
});
}
Minified (my first preference):
if(!String.prototype.replaceAll){Object.defineProperty(String.prototype,'replaceAll',{configurable:!0,writable:!0,enumerable:!1,value:function(search,replace){return this.replace(Object.prototype.toString.call(search)==='[object RegExp]'?search.global?search:RegExp(search.source,/\/([a-z]*)$/.exec(search.toString())[1]+'g'):RegExp(String(search).replace(/[.^$*+?()[{|\\]/g,"\\$&"),"g"),replace)}})}
Try it:
if(!String.prototype.replaceAll){Object.defineProperty(String.prototype,'replaceAll',{configurable:!0,writable:!0,enumerable:!1,value:function(search,replace){return this.replace(Object.prototype.toString.call(search)==='[object RegExp]'?search.global?search:RegExp(search.source,/\/([a-z]*)$/.exec(search.toString())[1]+'g'):RegExp(String(search).replace(/[.^$*+?()[{|\\]/g,"\\$&"),"g"),replace)}})}
console.log(
'Change this and this for me'.replaceAll('this','that')
); // Change that and that for me
console.log(
'aaaaaa'.replaceAll('aa','a')
); // aaa
console.log(
'{} (*) (*) (RegEx) (*) (\*) (\\*) [reserved characters]'.replaceAll('(*)','X')
); // {} X X (RegEx) X X (\*) [reserved characters]
console.log(
'How (replace) (XX) with $1?'.replaceAll(/(xx)/gi,'$$1')
); // How (replace) ($1) with $1?
console.log(
'Here is some numbers 1234567890 1000000 123123.'.replaceAll(/\d+/g,'***')
); // Here is some numbers *** *** *** and need to be replaced.
console.log(
'Remove numbers under 233: 236 229 711 200 5'.replaceAll(/\d+/g, function(m) {
return parseFloat(m) < 233 ? '' : m
})
); // Remove numbers under 233: 236 711
console.log(
'null'.replaceAll(null,'x')
); // x
// The difference between My first preference and the original:
// Now in 2022 with browsers > 2020 it should throw an error (But possible it be changed in future)
// console.log(
// 'xyz ABC abc ABC abc xyz'.replaceAll(/abc/i,'')
// );
// Browsers < 2020:
// xyz xyz
// Browsers > 2020
// TypeError: String.prototype.replaceAll called with a non-global RegExp
Browser support:
Internet Explorer 9 and later (rested on Internet Explorer 11).
All other browsers (after 2012).
The result is the same as the native replaceAll in case of the first argument input is:
null, undefined, Object, Function, Date, ... , RegExp, Number, String, ...
Ref: 22.1.3.19 String.prototype.replaceAll ( searchValue, replaceValue)
+ RegExp Syntax
Important note: As some professionals mention it, many of recursive functions that suggested in answers, will return the wrong result. (Try them with the challenged case of the above snippet.)
Maybe some tricky methods like .split('searchValue').join('replaceValue') or some well managed functions give same result, but definitely with much lower performance than native replaceAll() / polyfill replaceAll() / replace() + RegExp
Other methods of polyfill assignment
Naive, but supports even older browsers (be better to avoid)
For example, we can support IE7+ too, by not using Object.defineProperty() and using my old naive assignment method:
if (!String.prototype.replaceAll) {
String.prototype.replaceAll = function(search, replace) { // <-- Naive method for assignment
// ... (Polyfill code Here)
}
}
And it should work well for basic uses on IE7+.
But as here #sebastian-simon explained about, that can make secondary problems in case of more advanced uses. E.g.:
for (var k in 'hi') console.log(k);
// 0
// 1
// replaceAll <-- ?
Fully trustable, but heavy
In fact, my suggested option is a little optimistic. Like we trusted the environment (browser and Node.js), it is definitely for around 2012-2021. Also it is a standard/famous one, so it does not require any special consideration.
But there can be even older browsers or some unexpected problems, and polyfills still can support and solve more possible environment problems. So in case we need the maximum support that is possible, we can use polyfill libraries like:
https://polyfill.io/
Specially for replaceAll:
<script src="https://polyfill.io/v3/polyfill.min.js?features=String.prototype.replaceAll"></script>
The simplest way to do this without using any regular expression is split and join, like the code here:
var str = "Test abc test test abc test test test abc test test abc";
console.log(str.split('abc').join(''));
var str = "ff ff f f a de def";
str = str.replace(/f/g,'');
alert(str);
http://jsfiddle.net/ANHR9/
while (str.indexOf('abc') !== -1)
{
str = str.replace('abc', '');
}
If the string contains a similar pattern like abccc, you can use this:
str.replace(/abc(\s|$)/g, "")
As of August 2020 there is a Stage 4 proposal to ECMAScript that adds the replaceAll method to String.
It's now supported in Chrome 85+, Edge 85+, Firefox 77+, Safari 13.1+.
The usage is the same as the replace method:
String.prototype.replaceAll(searchValue, replaceValue)
Here's an example usage:
'Test abc test test abc test.'.replaceAll('abc', 'foo'); // -> 'Test foo test test foo test.'
It's supported in most modern browsers, but there exist polyfills:
core-js
es-shims
It is supported in the V8 engine behind an experimental flag --harmony-string-replaceall.
Read more on the V8 website.
The previous answers are way too complicated. Just use the replace function like this:
str.replace(/your_regex_pattern/g, replacement_string);
Example:
var str = "Test abc test test abc test test test abc test test abc";
var res = str.replace(/[abc]+/g, "");
console.log(res);
After several trials and a lot of fails, I found that the below function seems to be the best all-rounder when it comes to browser compatibility and ease of use. This is the only working solution for older browsers that I found. (Yes, even though old browser are discouraged and outdated, some legacy applications still make heavy use of OLE browsers (such as old Visual Basic 6 applications or Excel .xlsm macros with forms.)
Anyway, here's the simple function.
function replaceAll(str, match, replacement){
return str.split(match).join(replacement);
}
If you are trying to ensure that the string you are looking for won't exist even after the replacement, you need to use a loop.
For example:
var str = 'test aabcbc';
str = str.replace(/abc/g, '');
When complete, you will still have 'test abc'!
The simplest loop to solve this would be:
var str = 'test aabcbc';
while (str != str.replace(/abc/g, '')){
str.replace(/abc/g, '');
}
But that runs the replacement twice for each cycle. Perhaps (at risk of being voted down) that can be combined for a slightly more efficient but less readable form:
var str = 'test aabcbc';
while (str != (str = str.replace(/abc/g, ''))){}
// alert(str); alerts 'test '!
This can be particularly useful when looking for duplicate strings.
For example, if we have 'a,,,b' and we wish to remove all duplicate commas.
[In that case, one could do .replace(/,+/g,','), but at some point the regex gets complex and slow enough to loop instead.]
Given a string:
s = "Test abc test test abc test test test abc test test abc";
This seems to only remove the first occurrence of abc in the string above:
s = s.replace('abc', '');
How do I replace all occurrences of it?
As of August 2020: Modern browsers have support for the String.replaceAll() method defined by the ECMAScript 2021 language specification.
For older/legacy browsers:
function escapeRegExp(string) {
return string.replace(/[.*+?^${}()|[\]\\]/g, '\\$&'); // $& means the whole matched string
}
function replaceAll(str, find, replace) {
return str.replace(new RegExp(escapeRegExp(find), 'g'), replace);
}
Here is how this answer evolved:
str = str.replace(/abc/g, '');
In response to comment "what's if 'abc' is passed as a variable?":
var find = 'abc';
var re = new RegExp(find, 'g');
str = str.replace(re, '');
In response to Click Upvote's comment, you could simplify it even more:
function replaceAll(str, find, replace) {
return str.replace(new RegExp(find, 'g'), replace);
}
Note: Regular expressions contain special (meta) characters, and as such it is dangerous to blindly pass an argument in the find function above without pre-processing it to escape those characters. This is covered in the Mozilla Developer Network's JavaScript Guide on Regular Expressions, where they present the following utility function (which has changed at least twice since this answer was originally written, so make sure to check the MDN site for potential updates):
function escapeRegExp(string) {
return string.replace(/[.*+?^${}()|[\]\\]/g, '\\$&'); // $& means the whole matched string
}
So in order to make the replaceAll() function above safer, it could be modified to the following if you also include escapeRegExp:
function replaceAll(str, find, replace) {
return str.replace(new RegExp(escapeRegExp(find), 'g'), replace);
}
For the sake of completeness, I got to thinking about which method I should use to do this. There are basically two ways to do this as suggested by the other answers on this page.
Note: In general, extending the built-in prototypes in JavaScript is generally not recommended. I am providing as extensions on the String prototype simply for purposes of illustration, showing different implementations of a hypothetical standard method on the String built-in prototype.
Regular Expression Based Implementation
String.prototype.replaceAll = function(search, replacement) {
var target = this;
return target.replace(new RegExp(search, 'g'), replacement);
};
Split and Join (Functional) Implementation
String.prototype.replaceAll = function(search, replacement) {
var target = this;
return target.split(search).join(replacement);
};
Not knowing too much about how regular expressions work behind the scenes in terms of efficiency, I tended to lean toward the split and join implementation in the past without thinking about performance. When I did wonder which was more efficient, and by what margin, I used it as an excuse to find out.
On my Chrome Windows 8 machine, the regular expression based implementation is the fastest, with the split and join implementation being 53% slower. Meaning the regular expressions are twice as fast for the lorem ipsum input I used.
Check out this benchmark running these two implementations against each other.
As noted in the comment below by #ThomasLeduc and others, there could be an issue with the regular expression-based implementation if search contains certain characters which are reserved as special characters in regular expressions. The implementation assumes that the caller will escape the string beforehand or will only pass strings that are without the characters in the table in Regular Expressions (MDN).
MDN also provides an implementation to escape our strings. It would be nice if this was also standardized as RegExp.escape(str), but alas, it does not exist:
function escapeRegExp(str) {
return str.replace(/[.*+?^${}()|[\]\\]/g, "\\$&"); // $& means the whole matched string
}
We could call escapeRegExp within our String.prototype.replaceAll implementation, however, I'm not sure how much this will affect the performance (potentially even for strings for which the escape is not needed, like all alphanumeric strings).
In the latest versions of most popular browsers, you can use replaceAll
as shown here:
let result = "1 abc 2 abc 3".replaceAll("abc", "xyz");
// `result` is "1 xyz 2 xyz 3"
But check Can I use or another compatibility table first to make sure the browsers you're targeting have added support for it first.
For Node.js and compatibility with older/non-current browsers:
Note: Don't use the following solution in performance critical code.
As an alternative to regular expressions for a simple literal string, you could use
str = "Test abc test test abc test...".split("abc").join("");
The general pattern is
str.split(search).join(replacement)
This used to be faster in some cases than using replaceAll and a regular expression, but that doesn't seem to be the case anymore in modern browsers.
Benchmark: https://jsben.ch/TZYzj
Conclusion:
If you have a performance-critical use case (e.g., processing hundreds of strings), use the regular expression method. But for most typical use cases, this is well worth not having to worry about special characters.
Here's a string prototype function based on the accepted answer:
String.prototype.replaceAll = function(find, replace) {
var str = this;
return str.replace(new RegExp(find, 'g'), replace);
};
If your find contains special characters then you need to escape them:
String.prototype.replaceAll = function(find, replace) {
var str = this;
return str.replace(new RegExp(find.replace(/[-\/\\^$*+?.()|[\]{}]/g, '\\$&'), 'g'), replace);
};
Fiddle: http://jsfiddle.net/cdbzL/
Use word boundaries (\b)
'a cat is not a caterpillar'.replace(/\bcat\b/gi,'dog');
//"a dog is not a caterpillar"
This is a simple regex that avoids replacing parts of words in most cases. However, a dash - is still considered a word boundary. So conditionals can be used in this case to avoid replacing strings like cool-cat:
'a cat is not a cool-cat'.replace(/\bcat\b/gi,'dog');//wrong
//"a dog is not a cool-dog" -- nips
'a cat is not a cool-cat'.replace(/(?:\b([^-]))cat(?:\b([^-]))/gi,'$1dog$2');
//"a dog is not a cool-cat"
Basically, this question is the same as the question here:
Replace " ' " with " '' " in JavaScript
Regexp isn't the only way to replace multiple occurrences of a substring, far from it. Think flexible, think split!
var newText = "the cat looks like a cat".split('cat').join('dog');
Alternatively, to prevent replacing word parts—which the approved answer will do, too! You can get around this issue using regular expressions that are, I admit, somewhat more complex and as an upshot of that, a tad slower, too:
var regText = "the cat looks like a cat".replace(/(?:(^|[^a-z]))(([^a-z]*)(?=cat)cat)(?![a-z])/gi,"$1dog");
The output is the same as the accepted answer, however, using the /cat/g expression on this string:
var oops = 'the cat looks like a cat, not a caterpillar or coolcat'.replace(/cat/g,'dog');
//returns "the dog looks like a dog, not a dogerpillar or cooldog" ??
Oops indeed, this probably isn't what you want. What is, then? IMHO, a regex that only replaces 'cat' conditionally (i.e., not part of a word), like so:
var caterpillar = 'the cat looks like a cat, not a caterpillar or coolcat'.replace(/(?:(^|[^a-z]))(([^a-z]*)(?=cat)cat)(?![a-z])/gi,"$1dog");
//return "the dog looks like a dog, not a caterpillar or coolcat"
My guess is, this meets your needs. It's not foolproof, of course, but it should be enough to get you started. I'd recommend reading some more on these pages. This'll prove useful in perfecting this expression to meet your specific needs.
RegExp (regular expression) object
Regular-Expressions.info
Here is an example of .replace used with a callback function. In this case, it dramatically simplifies the expression and provides even more flexibility, like replacing with correct capitalisation or replacing both cat and cats in one go:
'Two cats are not 1 Cat! They\'re just cool-cats, you caterpillar'
.replace(/(^|.\b)(cat)(s?\b.|$)/gi,function(all,char1,cat,char2)
{
// Check 1st, capitalize if required
var replacement = (cat.charAt(0) === 'C' ? 'D' : 'd') + 'og';
if (char1 === ' ' && char2 === 's')
{ // Replace plurals, too
cat = replacement + 's';
}
else
{ // Do not replace if dashes are matched
cat = char1 === '-' || char2 === '-' ? cat : replacement;
}
return char1 + cat + char2;//return replacement string
});
//returns:
//Two dogs are not 1 Dog! They're just cool-cats, you caterpillar
These are the most common and readable methods.
var str = "Test abc test test abc test test test abc test test abc"
Method 1:
str = str.replace(/abc/g, "replaced text");
Method 2:
str = str.split("abc").join("replaced text");
Method 3:
str = str.replace(new RegExp("abc", "g"), "replaced text");
Method 4:
while(str.includes("abc")){
str = str.replace("abc", "replaced text");
}
Output:
console.log(str);
// Test replaced text test test replaced text test test test replaced text test test replaced text
Match against a global regular expression:
anotherString = someString.replace(/cat/g, 'dog');
For replacing a single time, use:
var res = str.replace('abc', "");
For replacing multiple times, use:
var res = str.replace(/abc/g, "");
str = str.replace(/abc/g, '');
Or try the replaceAll method, as recommended in this answer:
str = str.replaceAll('abc', '');
or:
var search = 'abc';
str = str.replaceAll(search, '');
EDIT: Clarification about replaceAll availability
The replaceAll method is added to String's prototype. This means it will be available for all string objects/literals.
Example:
var output = "test this".replaceAll('this', 'that'); // output is 'test that'.
output = output.replaceAll('that', 'this'); // output is 'test this'
Using RegExp in JavaScript could do the job for you. Just simply do something like below code, and don't forget the /g after which standout for global:
var str ="Test abc test test abc test test test abc test test abc";
str = str.replace(/abc/g, '');
If you think of reuse, create a function to do that for you, but it's not recommended as it's only one line function. But again, if you heavily use this, you can write something like this:
String.prototype.replaceAll = String.prototype.replaceAll || function(string, replaced) {
return this.replace(new RegExp(string, 'g'), replaced);
};
And simply use it in your code over and over like below:
var str ="Test abc test test abc test test test abc test test abc";
str = str.replaceAll('abc', '');
But as I mention earlier, it won't make a huge difference in terms of lines to be written or performance. Only caching the function may affect some faster performance on long strings and is also a good practice of DRY code if you want to reuse.
Say you want to replace all the 'abc' with 'x':
let some_str = 'abc def def lom abc abc def'.split('abc').join('x')
console.log(some_str) //x def def lom x x def
I was trying to think about something more simple than modifying the string prototype.
Use a regular expression:
str.replace(/abc/g, '');
Performance
Today 27.12.2019 I perform tests on macOS v10.13.6 (High Sierra) for the chosen solutions.
Conclusions
The str.replace(/abc/g, ''); (C) is a good cross-browser fast solution for all strings.
Solutions based on split-join (A,B) or replace (C,D) are fast
Solutions based on while (E,F,G,H) are slow - usually ~4 times slower for small strings and about ~3000 times (!) slower for long strings
The recurrence solutions (RA,RB) are slow and do not work for long strings
I also create my own solution. It looks like currently it is the shortest one which does the question job:
str.split`abc`.join``
str = "Test abc test test abc test test test abc test test abc";
str = str.split`abc`.join``
console.log(str);
Details
The tests were performed on Chrome 79.0, Safari 13.0.4 and Firefox 71.0 (64 bit). The tests RA and RB use recursion. Results
Short string - 55 characters
You can run tests on your machine HERE. Results for Chrome:
Long string: 275 000 characters
The recursive solutions RA and RB gives
RangeError: Maximum call stack size exceeded
For 1M characters they even break Chrome
I try to perform tests for 1M characters for other solutions, but E,F,G,H takes so much time that browser ask me to break script so I shrink test string to 275K characters. You can run tests on your machine HERE. Results for Chrome
Code used in tests
var t="Test abc test test abc test test test abc test test abc"; // .repeat(5000)
var log = (version,result) => console.log(`${version}: ${result}`);
function A(str) {
return str.split('abc').join('');
}
function B(str) {
return str.split`abc`.join``; // my proposition
}
function C(str) {
return str.replace(/abc/g, '');
}
function D(str) {
return str.replace(new RegExp("abc", "g"), '');
}
function E(str) {
while (str.indexOf('abc') !== -1) { str = str.replace('abc', ''); }
return str;
}
function F(str) {
while (str.indexOf('abc') !== -1) { str = str.replace(/abc/, ''); }
return str;
}
function G(str) {
while(str.includes("abc")) { str = str.replace('abc', ''); }
return str;
}
// src: https://stackoverflow.com/a/56989553/860099
function H(str)
{
let i = -1
let find = 'abc';
let newToken = '';
if (!str)
{
if ((str == null) && (find == null)) return newToken;
return str;
}
while ((
i = str.indexOf(
find, i >= 0 ? i + newToken.length : 0
)) !== -1
)
{
str = str.substring(0, i) +
newToken +
str.substring(i + find.length);
}
return str;
}
// src: https://stackoverflow.com/a/22870785/860099
function RA(string, prevstring) {
var omit = 'abc';
var place = '';
if (prevstring && string === prevstring)
return string;
prevstring = string.replace(omit, place);
return RA(prevstring, string)
}
// src: https://stackoverflow.com/a/26107132/860099
function RB(str) {
var find = 'abc';
var replace = '';
var i = str.indexOf(find);
if (i > -1){
str = str.replace(find, replace);
i = i + replace.length;
var st2 = str.substring(i);
if(st2.indexOf(find) > -1){
str = str.substring(0,i) + RB(st2, find, replace);
}
}
return str;
}
log('A ', A(t));
log('B ', B(t));
log('C ', C(t));
log('D ', D(t));
log('E ', E(t));
log('F ', F(t));
log('G ', G(t));
log('H ', H(t));
log('RA', RA(t)); // use reccurence
log('RB', RB(t)); // use reccurence
<p style="color:red">This snippet only presents codes used in tests. It not perform test itself!<p>
Replacing single quotes:
function JavaScriptEncode(text){
text = text.replace(/'/g,''')
// More encode here if required
return text;
}
Using
str = str.replace(new RegExp("abc", 'g'), "");
worked better for me than the previous answers. So new RegExp("abc", 'g') creates a regular expression what matches all occurrences ('g' flag) of the text ("abc"). The second part is what gets replaced to, in your case empty string ("").
str is the string, and we have to override it, as replace(...) just returns result, but not overrides. In some cases you might want to use that.
This is the fastest version that doesn't use regular expressions.
Revised jsperf
replaceAll = function(string, omit, place, prevstring) {
if (prevstring && string === prevstring)
return string;
prevstring = string.replace(omit, place);
return replaceAll(prevstring, omit, place, string)
}
It is almost twice as fast as the split and join method.
As pointed out in a comment here, this will not work if your omit variable contains place, as in: replaceAll("string", "s", "ss"), because it will always be able to replace another occurrence of the word.
There is another jsperf with variants on my recursive replace that go even faster (http://jsperf.com/replace-all-vs-split-join/12)!
Update July 27th 2017: It looks like RegExp now has the fastest performance in the recently released Chrome 59.
Loop it until number occurrences comes to 0, like this:
function replaceAll(find, replace, str) {
while (str.indexOf(find) > -1) {
str = str.replace(find, replace);
}
return str;
}
If what you want to find is already in a string, and you don't have a regex escaper handy, you can use join/split:
function replaceMulti(haystack, needle, replacement)
{
return haystack.split(needle).join(replacement);
}
someString = 'the cat looks like a cat';
console.log(replaceMulti(someString, 'cat', 'dog'));
function replaceAll(str, find, replace) {
var i = str.indexOf(find);
if (i > -1){
str = str.replace(find, replace);
i = i + replace.length;
var st2 = str.substring(i);
if(st2.indexOf(find) > -1){
str = str.substring(0,i) + replaceAll(st2, find, replace);
}
}
return str;
}
I like this method (it looks a little cleaner):
text = text.replace(new RegExp("cat","g"), "dog");
String.prototype.replaceAll - ECMAScript 2021
The new String.prototype.replaceAll() method returns a new string with all matches of a pattern replaced by a replacement. The pattern can be either a string or a RegExp, and the replacement can be either a string or a function to be called for each match.
const message = 'dog barks meow meow';
const messageFormatted = message.replaceAll('meow', 'woof')
console.log(messageFormatted);
Of course in 2021 the right answer is:
String.prototype.replaceAll()
console.log(
'Change this and this for me'.replaceAll('this','that') // Normal case
);
console.log(
'aaaaaa'.replaceAll('aa','a') // Challenged case
);
If you don't want to deal with replace() + RegExp.
But what if the browser is from before 2020?
In this case we need polyfill (forcing older browsers to support new features) (I think for a few years will be necessary).
I could not find a completely right method in answers. So I suggest this function that will be defined as a polyfill.
My suggested options for replaceAll polyfill:
replaceAll polyfill (with global-flag error) (more principled version)
if (!String.prototype.replaceAll) { // Check if the native function not exist
Object.defineProperty(String.prototype, 'replaceAll', { // Define replaceAll as a prototype for (Mother/Any) String
configurable: true, writable: true, enumerable: false, // Editable & non-enumerable property (As it should be)
value: function(search, replace) { // Set the function by closest input names (For good info in consoles)
return this.replace( // Using native String.prototype.replace()
Object.prototype.toString.call(search) === '[object RegExp]' // IsRegExp?
? search.global // Is the RegEx global?
? search // So pass it
: function(){throw new TypeError('replaceAll called with a non-global RegExp argument')}() // If not throw an error
: RegExp(String(search).replace(/[.^$*+?()[{|\\]/g, "\\$&"), "g"), // Replace all reserved characters with '\' then make a global 'g' RegExp
replace); // passing second argument
}
});
}
replaceAll polyfill (With handling global-flag missing by itself) (my first preference) - Why?
if (!String.prototype.replaceAll) { // Check if the native function not exist
Object.defineProperty(String.prototype, 'replaceAll', { // Define replaceAll as a prototype for (Mother/Any) String
configurable: true, writable: true, enumerable: false, // Editable & non-enumerable property (As it should be)
value: function(search, replace) { // Set the function by closest input names (For good info in consoles)
return this.replace( // Using native String.prototype.replace()
Object.prototype.toString.call(search) === '[object RegExp]' // IsRegExp?
? search.global // Is the RegEx global?
? search // So pass it
: RegExp(search.source, /\/([a-z]*)$/.exec(search.toString())[1] + 'g') // If not, make a global clone from the RegEx
: RegExp(String(search).replace(/[.^$*+?()[{|\\]/g, "\\$&"), "g"), // Replace all reserved characters with '\' then make a global 'g' RegExp
replace); // passing second argument
}
});
}
Minified (my first preference):
if(!String.prototype.replaceAll){Object.defineProperty(String.prototype,'replaceAll',{configurable:!0,writable:!0,enumerable:!1,value:function(search,replace){return this.replace(Object.prototype.toString.call(search)==='[object RegExp]'?search.global?search:RegExp(search.source,/\/([a-z]*)$/.exec(search.toString())[1]+'g'):RegExp(String(search).replace(/[.^$*+?()[{|\\]/g,"\\$&"),"g"),replace)}})}
Try it:
if(!String.prototype.replaceAll){Object.defineProperty(String.prototype,'replaceAll',{configurable:!0,writable:!0,enumerable:!1,value:function(search,replace){return this.replace(Object.prototype.toString.call(search)==='[object RegExp]'?search.global?search:RegExp(search.source,/\/([a-z]*)$/.exec(search.toString())[1]+'g'):RegExp(String(search).replace(/[.^$*+?()[{|\\]/g,"\\$&"),"g"),replace)}})}
console.log(
'Change this and this for me'.replaceAll('this','that')
); // Change that and that for me
console.log(
'aaaaaa'.replaceAll('aa','a')
); // aaa
console.log(
'{} (*) (*) (RegEx) (*) (\*) (\\*) [reserved characters]'.replaceAll('(*)','X')
); // {} X X (RegEx) X X (\*) [reserved characters]
console.log(
'How (replace) (XX) with $1?'.replaceAll(/(xx)/gi,'$$1')
); // How (replace) ($1) with $1?
console.log(
'Here is some numbers 1234567890 1000000 123123.'.replaceAll(/\d+/g,'***')
); // Here is some numbers *** *** *** and need to be replaced.
console.log(
'Remove numbers under 233: 236 229 711 200 5'.replaceAll(/\d+/g, function(m) {
return parseFloat(m) < 233 ? '' : m
})
); // Remove numbers under 233: 236 711
console.log(
'null'.replaceAll(null,'x')
); // x
// The difference between My first preference and the original:
// Now in 2022 with browsers > 2020 it should throw an error (But possible it be changed in future)
// console.log(
// 'xyz ABC abc ABC abc xyz'.replaceAll(/abc/i,'')
// );
// Browsers < 2020:
// xyz xyz
// Browsers > 2020
// TypeError: String.prototype.replaceAll called with a non-global RegExp
Browser support:
Internet Explorer 9 and later (rested on Internet Explorer 11).
All other browsers (after 2012).
The result is the same as the native replaceAll in case of the first argument input is:
null, undefined, Object, Function, Date, ... , RegExp, Number, String, ...
Ref: 22.1.3.19 String.prototype.replaceAll ( searchValue, replaceValue)
+ RegExp Syntax
Important note: As some professionals mention it, many of recursive functions that suggested in answers, will return the wrong result. (Try them with the challenged case of the above snippet.)
Maybe some tricky methods like .split('searchValue').join('replaceValue') or some well managed functions give same result, but definitely with much lower performance than native replaceAll() / polyfill replaceAll() / replace() + RegExp
Other methods of polyfill assignment
Naive, but supports even older browsers (be better to avoid)
For example, we can support IE7+ too, by not using Object.defineProperty() and using my old naive assignment method:
if (!String.prototype.replaceAll) {
String.prototype.replaceAll = function(search, replace) { // <-- Naive method for assignment
// ... (Polyfill code Here)
}
}
And it should work well for basic uses on IE7+.
But as here #sebastian-simon explained about, that can make secondary problems in case of more advanced uses. E.g.:
for (var k in 'hi') console.log(k);
// 0
// 1
// replaceAll <-- ?
Fully trustable, but heavy
In fact, my suggested option is a little optimistic. Like we trusted the environment (browser and Node.js), it is definitely for around 2012-2021. Also it is a standard/famous one, so it does not require any special consideration.
But there can be even older browsers or some unexpected problems, and polyfills still can support and solve more possible environment problems. So in case we need the maximum support that is possible, we can use polyfill libraries like:
https://polyfill.io/
Specially for replaceAll:
<script src="https://polyfill.io/v3/polyfill.min.js?features=String.prototype.replaceAll"></script>
The simplest way to do this without using any regular expression is split and join, like the code here:
var str = "Test abc test test abc test test test abc test test abc";
console.log(str.split('abc').join(''));
var str = "ff ff f f a de def";
str = str.replace(/f/g,'');
alert(str);
http://jsfiddle.net/ANHR9/
while (str.indexOf('abc') !== -1)
{
str = str.replace('abc', '');
}
If the string contains a similar pattern like abccc, you can use this:
str.replace(/abc(\s|$)/g, "")
As of August 2020 there is a Stage 4 proposal to ECMAScript that adds the replaceAll method to String.
It's now supported in Chrome 85+, Edge 85+, Firefox 77+, Safari 13.1+.
The usage is the same as the replace method:
String.prototype.replaceAll(searchValue, replaceValue)
Here's an example usage:
'Test abc test test abc test.'.replaceAll('abc', 'foo'); // -> 'Test foo test test foo test.'
It's supported in most modern browsers, but there exist polyfills:
core-js
es-shims
It is supported in the V8 engine behind an experimental flag --harmony-string-replaceall.
Read more on the V8 website.
The previous answers are way too complicated. Just use the replace function like this:
str.replace(/your_regex_pattern/g, replacement_string);
Example:
var str = "Test abc test test abc test test test abc test test abc";
var res = str.replace(/[abc]+/g, "");
console.log(res);
After several trials and a lot of fails, I found that the below function seems to be the best all-rounder when it comes to browser compatibility and ease of use. This is the only working solution for older browsers that I found. (Yes, even though old browser are discouraged and outdated, some legacy applications still make heavy use of OLE browsers (such as old Visual Basic 6 applications or Excel .xlsm macros with forms.)
Anyway, here's the simple function.
function replaceAll(str, match, replacement){
return str.split(match).join(replacement);
}
If you are trying to ensure that the string you are looking for won't exist even after the replacement, you need to use a loop.
For example:
var str = 'test aabcbc';
str = str.replace(/abc/g, '');
When complete, you will still have 'test abc'!
The simplest loop to solve this would be:
var str = 'test aabcbc';
while (str != str.replace(/abc/g, '')){
str.replace(/abc/g, '');
}
But that runs the replacement twice for each cycle. Perhaps (at risk of being voted down) that can be combined for a slightly more efficient but less readable form:
var str = 'test aabcbc';
while (str != (str = str.replace(/abc/g, ''))){}
// alert(str); alerts 'test '!
This can be particularly useful when looking for duplicate strings.
For example, if we have 'a,,,b' and we wish to remove all duplicate commas.
[In that case, one could do .replace(/,+/g,','), but at some point the regex gets complex and slow enough to loop instead.]
Probably a simple one but my knowledge of creating regular expressions is a little vague.
I'm trying to match any string followed by a comma and a space except if it is 'Bair Hugger' or 'Fluid Warmer'
Here is what I have so far
var re_comma = new RegExp("\w+[^Bair Hugger|Fluid Warmer]" + ", ", "i");
Any ideas?
New answer
Regarding your example I'd say it is really easier to split the string and iterate over it:
function filter(str, delim, test) {
var parts = str.split(delim),
result = [];
for(var i = 0, len = parts.length; i < len; i++) {
if(test(parts[i])) result.push(parts[i]);
}
return result.join(delim);
}
str = filter(str, ', ', function(s) {
s = s.toLowerCase();
return s === 'bair hugger' || s === 'fluid warmer';
});
Otherwise, your expression becomes something like this:
new RegExp("(^|, )(?!(?:Bair Hugger|Fluid Warmer)(?:$|, )).+?(, |$)", "i");
and you have to use a callback for the replacement to decide whether to remove the preceding , or trailing , or not:
str = str.replace(re_comma, function(str, pre, tail) {
return pre && tail ? tail : '';// middle of the string, leave one
});
The intention of this code is less clear. Maybe there is a simpler expression, but I think filtering the array is still cleaner.
Old answer: (doesn't solve the problem at hand but provides information regarding regular expressions).
[] denotes a character class and will only match one character out of the ones you provided. [^Bair Hugger|Fluid Warmer] is the same as [^Bair Huge|FldWm].
You could use a negative lookahead:
new RegExp("^(?!(Bair Hugger|Fluid Warmer), ).+?, $", "i");
Note that you have to use \\ inside a string to produce one \. Otherwise, "\w" becomes w and is not a special character sequence anymore.You also have to anchor the expression.
Update: As you mentioned you want to match any string before the comma, I decided to use . instead of \w, to match any character.
I love regex and Felix Kling answer is correct :)
However for such simple matching I would normally use something like below
function contains(str, text) {
return str.indexOf(text) >= 0;
}
if(contains(myString, 'random')) {
//myString contains "random"
}
Solution:
reg =/(?:Bair Hugger|Fluid Warmer),| (.*)/
str='Bair Hugger, lalala'
reg.exec(str)
>> ["Bair Hugger, lalala", " lalala"]
newStr = reg.exec(str)[1]
Is it possible to do something like this?
var pattern = /some regex segment/ + /* comment here */
/another segment/;
Or do I have to use new RegExp() syntax and concatenate a string? I'd prefer to use the literal as the code is both more self-evident and concise.
Here is how to create a regular expression without using the regular expression literal syntax. This lets you do arbitary string manipulation before it becomes a regular expression object:
var segment_part = "some bit of the regexp";
var pattern = new RegExp("some regex segment" + /*comment here */
segment_part + /* that was defined just now */
"another segment");
If you have two regular expression literals, you can in fact concatenate them using this technique:
var regex1 = /foo/g;
var regex2 = /bar/y;
var flags = (regex1.flags + regex2.flags).split("").sort().join("").replace(/(.)(?=.*\1)/g, "");
var regex3 = new RegExp(expression_one.source + expression_two.source, flags);
// regex3 is now /foobar/gy
It's just more wordy than just having expression one and two being literal strings instead of literal regular expressions.
Just randomly concatenating regular expressions objects can have some adverse side effects. Use the RegExp.source instead:
var r1 = /abc/g;
var r2 = /def/;
var r3 = new RegExp(r1.source + r2.source,
(r1.global ? 'g' : '')
+ (r1.ignoreCase ? 'i' : '') +
(r1.multiline ? 'm' : ''));
console.log(r3);
var m = 'test that abcdef and abcdef has a match?'.match(r3);
console.log(m);
// m should contain 2 matches
This will also give you the ability to retain the regular expression flags from a previous RegExp using the standard RegExp flags.
jsFiddle
I don't quite agree with the "eval" option.
var xxx = /abcd/;
var yyy = /efgh/;
var zzz = new RegExp(eval(xxx)+eval(yyy));
will give "//abcd//efgh//" which is not the intended result.
Using source like
var zzz = new RegExp(xxx.source+yyy.source);
will give "/abcdefgh/" and that is correct.
Logicaly there is no need to EVALUATE, you know your EXPRESSION. You just need its SOURCE or how it is written not necessarely its value. As for the flags, you just need to use the optional argument of RegExp.
In my situation, I do run in the issue of ^ and $ being used in several expression I am trying to concatenate together! Those expressions are grammar filters used accross the program. Now I wan't to use some of them together to handle the case of PREPOSITIONS.
I may have to "slice" the sources to remove the starting and ending ^( and/or )$ :)
Cheers, Alex.
Problem If the regexp contains back-matching groups like \1.
var r = /(a|b)\1/ // Matches aa, bb but nothing else.
var p = /(c|d)\1/ // Matches cc, dd but nothing else.
Then just contatenating the sources will not work. Indeed, the combination of the two is:
var rp = /(a|b)\1(c|d)\1/
rp.test("aadd") // Returns false
The solution:
First we count the number of matching groups in the first regex, Then for each back-matching token in the second, we increment it by the number of matching groups.
function concatenate(r1, r2) {
var count = function(r, str) {
return str.match(r).length;
}
var numberGroups = /([^\\]|^)(?=\((?!\?:))/g; // Home-made regexp to count groups.
var offset = count(numberGroups, r1.source);
var escapedMatch = /[\\](?:(\d+)|.)/g; // Home-made regexp for escaped literals, greedy on numbers.
var r2newSource = r2.source.replace(escapedMatch, function(match, number) { return number?"\\"+(number-0+offset):match; });
return new RegExp(r1.source+r2newSource,
(r1.global ? 'g' : '')
+ (r1.ignoreCase ? 'i' : '')
+ (r1.multiline ? 'm' : ''));
}
Test:
var rp = concatenate(r, p) // returns /(a|b)\1(c|d)\2/
rp.test("aadd") // Returns true
Providing that:
you know what you do in your regexp;
you have many regex pieces to form a pattern and they will use same flag;
you find it more readable to separate your small pattern chunks into an array;
you also want to be able to comment each part for next dev or yourself later;
you prefer to visually simplify your regex like /this/g rather than new RegExp('this', 'g');
it's ok for you to assemble the regex in an extra step rather than having it in one piece from the start;
Then you may like to write this way:
var regexParts =
[
/\b(\d+|null)\b/,// Some comments.
/\b(true|false)\b/,
/\b(new|getElementsBy(?:Tag|Class|)Name|arguments|getElementById|if|else|do|null|return|case|default|function|typeof|undefined|instanceof|this|document|window|while|for|switch|in|break|continue|length|var|(?:clear|set)(?:Timeout|Interval))(?=\W)/,
/(\$|jQuery)/,
/many more patterns/
],
regexString = regexParts.map(function(x){return x.source}).join('|'),
regexPattern = new RegExp(regexString, 'g');
you can then do something like:
string.replace(regexPattern, function()
{
var m = arguments,
Class = '';
switch(true)
{
// Numbers and 'null'.
case (Boolean)(m[1]):
m = m[1];
Class = 'number';
break;
// True or False.
case (Boolean)(m[2]):
m = m[2];
Class = 'bool';
break;
// True or False.
case (Boolean)(m[3]):
m = m[3];
Class = 'keyword';
break;
// $ or 'jQuery'.
case (Boolean)(m[4]):
m = m[4];
Class = 'dollar';
break;
// More cases...
}
return '<span class="' + Class + '">' + m + '</span>';
})
In my particular case (a code-mirror-like editor), it is much easier to perform one big regex, rather than a lot of replaces like following as each time I replace with a html tag to wrap an expression, the next pattern will be harder to target without affecting the html tag itself (and without the good lookbehind that is unfortunately not supported in javascript):
.replace(/(\b\d+|null\b)/g, '<span class="number">$1</span>')
.replace(/(\btrue|false\b)/g, '<span class="bool">$1</span>')
.replace(/\b(new|getElementsBy(?:Tag|Class|)Name|arguments|getElementById|if|else|do|null|return|case|default|function|typeof|undefined|instanceof|this|document|window|while|for|switch|in|break|continue|var|(?:clear|set)(?:Timeout|Interval))(?=\W)/g, '<span class="keyword">$1</span>')
.replace(/\$/g, '<span class="dollar">$</span>')
.replace(/([\[\](){}.:;,+\-?=])/g, '<span class="ponctuation">$1</span>')
It would be preferable to use the literal syntax as often as possible. It's shorter, more legible, and you do not need escape quotes or double-escape backlashes. From "Javascript Patterns", Stoyan Stefanov 2010.
But using New may be the only way to concatenate.
I would avoid eval. Its not safe.
You could do something like:
function concatRegex(...segments) {
return new RegExp(segments.join(''));
}
The segments would be strings (rather than regex literals) passed in as separate arguments.
You can concat regex source from both the literal and RegExp class:
var xxx = new RegExp(/abcd/);
var zzz = new RegExp(xxx.source + /efgh/.source);
Use the constructor with 2 params and avoid the problem with trailing '/':
var re_final = new RegExp("\\" + ".", "g"); // constructor can have 2 params!
console.log("...finally".replace(re_final, "!") + "\n" + re_final +
" works as expected..."); // !!!finally works as expected
// meanwhile
re_final = new RegExp("\\" + "." + "g"); // appends final '/'
console.log("... finally".replace(re_final, "!")); // ...finally
console.log(re_final, "does not work!"); // does not work
No, the literal way is not supported. You'll have to use RegExp.
the easier way to me would be concatenate the sources, ex.:
a = /\d+/
b = /\w+/
c = new RegExp(a.source + b.source)
the c value will result in:
/\d+\w+/
I prefer to use eval('your expression') because it does not add the /on each end/ that ='new RegExp' does.