Related
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.]
How do I go about assigning certain words a unique id tag using vanilla javascript? For example:
<p>All the king's horses ran away.</p>
The word "All" gets the id "term1", "King's" gets "term2", "away" gets "term3", etc. but not every word in the sentence will get assigned an id.
I am currently using the replace method but I think it's the wrong approach:
var str = document.getElementsByTagName('p')[0].innerHTML;
function addId() {
txt = str.replace(/all/i, '<span id="term1">$&</span>').replace(/king's/i, '<span id="term2">$&</span>').replace(/away/i, '<span id="term3">$&</span>');
document.getElementsByTagName('p')[0].innerHTML = txt;
}
window.onload = function() {
addId();
};
<p>All the king's horses ran away.</p>
This forces me to chain a bunch of replace commands, changing the id name each time. I don't think this is the best solution. What is best way to do this? Thanks for your help!
I believe this should make the job done. You can replace blacklist with whitelist. That will depend on your use case. Also, addIds can use Array.map instead of Array.forEach that will make the whole function a one-liner. This example is imperative because it will be more readable.
// string - string where we want to add ids
// blackList - words we want to skip (can be whiteliste black list is more general)
function addIds(string, blackList = ['the', 'a']) {
const stringArray = string.split(' ') // split string into separate words
let stringWithIds = string // this will be final string
stringArray.forEach((item, index) => {
// skip word if black listed
if (blackList.indexOf(item.toLowerCase()) !== -1) {
return
}
stringWithIds = stringWithIds.replace(item, `<span id="term${index}">${item}</span>`) // add id to word if not black listed
})
// replace string with our string with ids
document.getElementsByTagName('p')[0].innerHTML = stringWithIds;
}
window.onload = function() {
const str = document.getElementsByTagName('p')[0].innerHTML;
addIds(str); // you can pass custom blacklist as optional second parameter
};
<p>All the king's horses ran away.</p>
I think this function would be flexible enough for the task:
// Here, your logic to decide if the word must be tagged
const shouldBeTagged = word => {
return true
}
function addId(str) {
// Split the string into individual words to deal with each of them
// one by one
let txt = str.split(' ').map((word, i) => {
// Should this word be tagged
if (shouldBeTagged(word)) {
// If so, tag it
return '<span id="term' + i + '">' + word + '</span>'
} else {
// Otherwise, return the naked word
return word
}
// Join the words together again
}).join(' ')
document.getElementsByTagName('p')[0].innerHTML = txt;
}
window.onload = function() {
addId(document.getElementsByTagName('p')[0].innerHTML);
};
<p>All the king's horses ran away.</p>
Given a string:
s = "Test abc test test abc test test test abc test test abc";
This seems to only remove the first occurrence of abc in the string above:
s = s.replace('abc', '');
How do I replace all occurrences of it?
As of August 2020: Modern browsers have support for the String.replaceAll() method defined by the ECMAScript 2021 language specification.
For older/legacy browsers:
function escapeRegExp(string) {
return string.replace(/[.*+?^${}()|[\]\\]/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.]
I'm trying to do something that would be similar to turning a url slug-like variable into text that could be used for a title.
So, I have a variable for example that is like this:
var thisID = 'athlete-profile';
function myFunc(thisID) {
// i need to use thisID as the id and href in a loop that generates a string of <li><a>'s\
function makeTitle(thisID) {
// convert thisID to text so for this example it would return 'Athlete Profile'
return 'Athlete Profile';
}
for () {
var str = '<li id="'+thisID+'">'+makeTitle(thisID)+'';
}
// make sense?
}
I'd like to not use a regex to do this if possible somehow, but I don't think there's a way to do it without one. So any one who knows how to do this type of thing let me know, it would be a great help.
Thanks
I would advise you to use regular expression. But if you really don't want to use regular expressions, the solution below would work for simple cases. Feel free to modify it as you like it.
function makeTitle(slug) {
var words = slug.split('-');
for (var i = 0; i < words.length; i++) {
var word = words[i];
words[i] = word.charAt(0).toUpperCase() + word.slice(1);
}
return words.join(' ');
}
console.log(
makeTitle("athlete-profile")
)
function titleize(slug) {
var words = slug.split("-");
return words.map(function(word) {
return word.charAt(0).toUpperCase() + word.substring(1).toLowerCase();
}).join(' ');
}
console.log(titleize("athlete-profile"))
It works pretty simply:
It splits the string by - into words.
It maps each word into title case.
It joins the resulting words with spaces.
Do it in one line:
'athlete-profile'.split("-").join(" ").replace(/\w\S*/g, function(txt){return txt.charAt(0).toUpperCase() + txt.substr(1).toLowerCase()})
Output: Athlete Profile
The makeTitle() part of your question can be implemented something like this:
function makeTitle(thisID) {
return thisID.replace(/-/g, " ").replace(/\b[a-z]/g, function() {
return arguments[0].toUpperCase();
});
}
console.log(makeTitle("athlete-profile"))
The first .replace() changes all hyphens to spaces, and then the second .replace() takes any lower-case letter that follows a word boundary and makes it upper-case.
(For more information see the MDN doco for .replace().)
As far as doing it without using regular expressions, I'm not sure why you'd specifically want to avoid them, especially when the required expressions are pretty simple in this case (especially if you do the hyphen to space and first letter capitalisation in two steps as shown above). But there are endless ways to do this without regex using various combinations of JavaScript's string manipulation methods.
Do it like this
let someString = 'im a string';
console.log(someString.replace(/-/g, ' ')
.replace(/\w\S*/g, function (txt) {
return
txt.charAt(0).toUpperCase() + txt.substr(1).toLowerCase()
})
)
Output: Im A String
Short and great way:
const slugToText = (slug) => {
return slug.toLowerCase().replace(/-/g,' ')
}
Much Simplified answer
we can use String.prototype.replaceAll method to easily achieve this
function convertSlugToString(slug) {
return slug.replaceAll("-", " ");
}
incase you want to make sure the output is all lowercase then you can do the following
function convertSlugToString(slug) {
return slug.toLowerCase().replaceAll("-", " ");
}
Additional info:
String.prototype.replaceAll() is a ES2021 feature and it also has a great browser support with 93.64% global coverage, click here for more info
if you want to support IE then refer to the other answers
First a quick definition :)
Template - A string which may contain placeholders (example:"hello [name]")
Placeholder - A substring whitin square brackets (example: "name" in "hello [name]:).
Properties map - A valid object with strings as values
I need to write a code that replace placeholders (along with brackets) with the matching values in the properties map.
example:
for the following properties map:
{
"name":"world",
"my":"beautiful",
"a":"[b]",
"b":"c",
"c":"my"
}
Expected results:
"hello name" -> "hello name"
"hello [name]" -> "hello world"
"[b]" -> "c"
"[a]" -> "c" (because [a]->[b]->[c])
"[[b]]" -> "my" (because [[b]]->[c]->my)
"hello [my] [name]" -> "hello beautiful world"
var map = {
"name":"world",
"my":"beautiful",
"a":"[b]",
"b":"c",
"c":"my"
};
var str = "hello [my] [name] [[b]]";
do {
var strBeforeReplace = str;
for (var k in map) {
if (!map.hasOwnProperty(k)) continue;
var needle = "[" + k + "]";
str = str.replace(needle, map[k]);
}
var strChanged = str !== strBeforeReplace;
} while (strChanged);
document.write(str); //hello beautiful world my
The answer by #chris is excellent, I just want to provide an alternative solution using regular expressions that works "the other way round", i.e., not by looking for occurrences of the "placeholder versions" of all items in the properties map, but by repeatedly looking for occurrences of the placeholder itself, and substituting it with the corresponding value from the property map. This has two advantages:
If the property map grows very large, this solution should have
better performance (still to be benchmarked though).
The placeholder and the way substitutions work can easily be modified by adjusting the regular expression and the substitution function (might not be an issue here).
The downside is, of course, that the code is a little more complex (partly due to the fact that JavaScript lacks a nice way of substituting regular expression matches using custom functions, so that's what substituteRegExp is for):
function substituteRegExp(string, regexp, f) {
// substitute all matches of regexp in string with the value
// returned by f given a match and the corresponding group values
var found;
var lastIndex = 0;
var result = "";
while (found = regexp.exec(string)) {
var subst = f.apply(this, found);
result += string.slice(lastIndex, found.index) + subst;
lastIndex = found.index + found[0].length;
}
result += string.slice(lastIndex);
return result;
}
function templateReplace(string, values) {
// repeatedly substitute [key] placeholders in string by values[key]
var placeholder = /\[([a-zA-Z0-9]+)\]/g;
while (true) {
var newString = substituteRegExp(string, placeholder, function(match, key) {
return values[key];
});
if (newString == string)
break;
string = newString;
}
return string;
}
alert(templateReplace("hello [[b]] [my] [name]", {
"name":"world",
"my":"beautiful",
"a":"[b]",
"b":"c",
"c":"my"
})); // -> "hello my beautiful world"
Update: I did some little profiling to compare the two solutions (jsFiddle at http://jsfiddle.net/n8Fyv/1/, I also used Firebug). While #chris' solution is faster for small strings (no need for parsing the regular expression etc), this solution performs a lot better for large strings (in the order of thousands of characters). I did not compare for different sizes of the property map, but expect even bigger differences there.
In theory, this solution has runtime O(k n) where k is the depth of nesting of placeholders and n is the length of the string (assuming dictionary/hash lookups need constant time), while #chris' solution is O(k n m) where m is the number of items in the property map. All of this is only relevant for large inputs, of course.
If you're familiar with .NET's String.Format, then you should take a look at this JavaScript implementation. It supports number formatting too, just like String.Format.
Here's an example of how to use it:
var result = String.Format("Hello {my} {name}", map);
However, it would require some modification to do recursive templates.