How to split a long regular expression into multiple lines in JavaScript? - javascript

I have a very long regular expression, which I wish to split into multiple lines in my JavaScript code to keep each line length 80 characters according to JSLint rules. It's just better for reading, I think.
Here's pattern sample:
var pattern = /^(([^<>()[\]\\.,;:\s#\"]+(\.[^<>()[\]\\.,;:\s#\"]+)*)|(\".+\"))#((\[[0-9]{1,3}\.[0-9]{1,3}\.[0-9]{1,3}\.[0-9]{1,3}\])|(([a-zA-Z\-0-9]+\.)+[a-zA-Z]{2,}))$/;

Extending #KooiInc answer, you can avoid manually escaping every special character by using the source property of the RegExp object.
Example:
var urlRegex= new RegExp(''
+ /(?:(?:(https?|ftp):)?\/\/)/.source // protocol
+ /(?:([^:\n\r]+):([^#\n\r]+)#)?/.source // user:pass
+ /(?:(?:www\.)?([^\/\n\r]+))/.source // domain
+ /(\/[^?\n\r]+)?/.source // request
+ /(\?[^#\n\r]*)?/.source // query
+ /(#?[^\n\r]*)?/.source // anchor
);
or if you want to avoid repeating the .source property you can do it using the Array.map() function:
var urlRegex= new RegExp([
/(?:(?:(https?|ftp):)?\/\/)/ // protocol
,/(?:([^:\n\r]+):([^#\n\r]+)#)?/ // user:pass
,/(?:(?:www\.)?([^\/\n\r]+))/ // domain
,/(\/[^?\n\r]+)?/ // request
,/(\?[^#\n\r]*)?/ // query
,/(#?[^\n\r]*)?/ // anchor
].map(function(r) {return r.source}).join(''));
In ES6 the map function can be reduced to:
.map(r => r.source)

[Edit 2022/08] Created a small github repository to create regular expressions with spaces, comments and templating.
You could convert it to a string and create the expression by calling new RegExp():
var myRE = new RegExp (['^(([^<>()[\]\\.,;:\\s#\"]+(\\.[^<>(),[\]\\.,;:\\s#\"]+)*)',
'|(\\".+\\"))#((\\[[0-9]{1,3}\\.[0-9]{1,3}\\.[0-9]{1,3}\\.',
'[0-9]{1,3}\])|(([a-zA-Z\-0-9]+\\.)+',
'[a-zA-Z]{2,}))$'].join(''));
Notes:
when converting the expression literal to a string you need to escape all backslashes as backslashes are consumed when evaluating a string literal. (See Kayo's comment for more detail.)
RegExp accepts modifiers as a second parameter
/regex/g => new RegExp('regex', 'g')
[Addition ES20xx (tagged template)]
In ES20xx you can use tagged templates. See the snippet.
Note:
Disadvantage here is that you can't use plain whitespace in the regular expression string (always use \s, \s+, \s{1,x}, \t, \n etc).
(() => {
const createRegExp = (str, opts) =>
new RegExp(str.raw[0].replace(/\s/gm, ""), opts || "");
const yourRE = createRegExp`
^(([^<>()[\]\\.,;:\s#\"]+(\.[^<>()[\]\\.,;:\s#\"]+)*)|
(\".+\"))#((\[[0-9]{1,3}\.[0-9]{1,3}\.[0-9]{1,3}\.[0-9]{1,3}\])|
(([a-zA-Z\-0-9]+\.)+[a-zA-Z]{2,}))$`;
console.log(yourRE);
const anotherLongRE = createRegExp`
(\byyyy\b)|(\bm\b)|(\bd\b)|(\bh\b)|(\bmi\b)|(\bs\b)|(\bms\b)|
(\bwd\b)|(\bmm\b)|(\bdd\b)|(\bhh\b)|(\bMI\b)|(\bS\b)|(\bMS\b)|
(\bM\b)|(\bMM\b)|(\bdow\b)|(\bDOW\b)
${"gi"}`;
console.log(anotherLongRE);
})();

Using strings in new RegExp is awkward because you must escape all the backslashes. You may write smaller regexes and concatenate them.
Let's split this regex
/^foo(.*)\bar$/
We will use a function to make things more beautiful later
function multilineRegExp(regs, options) {
return new RegExp(regs.map(
function(reg){ return reg.source; }
).join(''), options);
}
And now let's rock
var r = multilineRegExp([
/^foo/, // we can add comments too
/(.*)/,
/\bar$/
]);
Since it has a cost, try to build the real regex just once and then use that.

Thanks to the wonderous world of template literals you can now write big, multi-line, well-commented, and even semantically nested regexes in ES6.
//build regexes without worrying about
// - double-backslashing
// - adding whitespace for readability
// - adding in comments
let clean = (piece) => (piece
.replace(/((^|\n)(?:[^\/\\]|\/[^*\/]|\\.)*?)\s*\/\*(?:[^*]|\*[^\/])*(\*\/|)/g, '$1')
.replace(/((^|\n)(?:[^\/\\]|\/[^\/]|\\.)*?)\s*\/\/[^\n]*/g, '$1')
.replace(/\n\s*/g, '')
);
window.regex = ({raw}, ...interpolations) => (
new RegExp(interpolations.reduce(
(regex, insert, index) => (regex + insert + clean(raw[index + 1])),
clean(raw[0])
))
);
Using this you can now write regexes like this:
let re = regex`I'm a special regex{3} //with a comment!`;
Outputs
/I'm a special regex{3}/
Or what about multiline?
'123hello'
.match(regex`
//so this is a regex
//here I am matching some numbers
(\d+)
//Oh! See how I didn't need to double backslash that \d?
([a-z]{1,3}) /*note to self, this is group #2*/
`)
[2]
Outputs hel, neat!
"What if I need to actually search a newline?", well then use \n silly!
Working on my Firefox and Chrome.
Okay, "how about something a little more complex?"
Sure, here's a piece of an object destructuring JS parser I was working on:
regex`^\s*
(
//closing the object
(\})|
//starting from open or comma you can...
(?:[,{]\s*)(?:
//have a rest operator
(\.\.\.)
|
//have a property key
(
//a non-negative integer
\b\d+\b
|
//any unencapsulated string of the following
\b[A-Za-z$_][\w$]*\b
|
//a quoted string
//this is #5!
("|')(?:
//that contains any non-escape, non-quote character
(?!\5|\\).
|
//or any escape sequence
(?:\\.)
//finished by the quote
)*\5
)
//after a property key, we can go inside
\s*(:|)
|
\s*(?={)
)
)
((?:
//after closing we expect either
// - the parent's comma/close,
// - or the end of the string
\s*(?:[,}\]=]|$)
|
//after the rest operator we expect the close
\s*\}
|
//after diving into a key we expect that object to open
\s*[{[:]
|
//otherwise we saw only a key, we now expect a comma or close
\s*[,}{]
).*)
$`
It outputs /^\s*((\})|(?:[,{]\s*)(?:(\.\.\.)|(\b\d+\b|\b[A-Za-z$_][\w$]*\b|("|')(?:(?!\5|\\).|(?:\\.))*\5)\s*(:|)|\s*(?={)))((?:\s*(?:[,}\]=]|$)|\s*\}|\s*[{[:]|\s*[,}{]).*)$/
And running it with a little demo?
let input = '{why, hello, there, "you huge \\"", 17, {big,smelly}}';
for (
let parsed;
parsed = input.match(r);
input = parsed[parsed.length - 1]
) console.log(parsed[1]);
Successfully outputs
{why
, hello
, there
, "you huge \""
, 17
,
{big
,smelly
}
}
Note the successful capturing of the quoted string.
I tested it on Chrome and Firefox, works a treat!
If curious you can checkout what I was doing, and its demonstration.
Though it only works on Chrome, because Firefox doesn't support backreferences or named groups. So note the example given in this answer is actually a neutered version and might get easily tricked into accepting invalid strings.

There are good answers here, but for completeness someone should mention Javascript's core feature of inheritance with the prototype chain. Something like this illustrates the idea:
RegExp.prototype.append = function(re) {
return new RegExp(this.source + re.source, this.flags);
};
let regex = /[a-z]/g
.append(/[A-Z]/)
.append(/[0-9]/);
console.log(regex); //=> /[a-z][A-Z][0-9]/g

The regex above is missing some black slashes which isn't working properly. So, I edited the regex. Please consider this regex which works 99.99% for email validation.
let EMAIL_REGEXP =
new RegExp (['^(([^<>()[\\]\\\.,;:\\s#\"]+(\\.[^<>()\\[\\]\\\.,;:\\s#\"]+)*)',
'|(".+"))#((\\[[0-9]{1,3}\\.[0-9]{1,3}\\.[0-9]{1,3}\\.',
'[0-9]{1,3}\])|(([a-zA-Z\\-0-9]+\\.)+',
'[a-zA-Z]{2,}))$'].join(''));

To avoid the Array join, you can also use the following syntax:
var pattern = new RegExp('^(([^<>()[\]\\.,;:\s#\"]+' +
'(\.[^<>()[\]\\.,;:\s#\"]+)*)|(\".+\"))#' +
'((\[[0-9]{1,3}\.[0-9]{1,3}\.[0-9]{1,3}\.[0-9]{1,3}\])|' +
'(([a-zA-Z\-0-9]+\.)+[a-zA-Z]{2,}))$');

You can simply use string operation.
var pattenString = "^(([^<>()[\]\\.,;:\s#\"]+(\.[^<>()[\]\\.,;:\s#\"]+)*)|"+
"(\".+\"))#((\[[0-9]{1,3}\.[0-9]{1,3}\.[0-9]{1,3}\.[0-9]{1,3}\])|"+
"(([a-zA-Z\-0-9]+\.)+[a-zA-Z]{2,}))$";
var patten = new RegExp(pattenString);

I tried improving korun's answer by encapsulating everything and implementing support for splitting capturing groups and character sets - making this method much more versatile.
To use this snippet you need to call the variadic function combineRegex whose arguments are the regular expression objects you need to combine. Its implementation can be found at the bottom.
Capturing groups can't be split directly that way though as it would leave some parts with just one parenthesis. Your browser would fail with an exception.
Instead I'm simply passing the contents of the capture group inside an array. The parentheses are automatically added when combineRegex encounters an array.
Furthermore quantifiers need to follow something. If for some reason the regular expression needs to be split in front of a quantifier you need to add a pair of parentheses. These will be removed automatically. The point is that an empty capture group is pretty useless and this way quantifiers have something to refer to. The same method can be used for things like non-capturing groups (/(?:abc)/ becomes [/()?:abc/]).
This is best explained using a simple example:
var regex = /abcd(efghi)+jkl/;
would become:
var regex = combineRegex(
/ab/,
/cd/,
[
/ef/,
/ghi/
],
/()+jkl/ // Note the added '()' in front of '+'
);
If you must split character sets you can use objects ({"":[regex1, regex2, ...]}) instead of arrays ([regex1, regex2, ...]). The key's content can be anything as long as the object only contains one key. Note that instead of () you have to use ] as dummy beginning if the first character could be interpreted as quantifier. I.e. /[+?]/ becomes {"":[/]+?/]}
Here is the snippet and a more complete example:
function combineRegexStr(dummy, ...regex)
{
return regex.map(r => {
if(Array.isArray(r))
return "("+combineRegexStr(dummy, ...r).replace(dummy, "")+")";
else if(Object.getPrototypeOf(r) === Object.getPrototypeOf({}))
return "["+combineRegexStr(/^\]/, ...(Object.entries(r)[0][1]))+"]";
else
return r.source.replace(dummy, "");
}).join("");
}
function combineRegex(...regex)
{
return new RegExp(combineRegexStr(/^\(\)/, ...regex));
}
//Usage:
//Original:
console.log(/abcd(?:ef[+A-Z0-9]gh)+$/.source);
//Same as:
console.log(
combineRegex(
/ab/,
/cd/,
[
/()?:ef/,
{"": [/]+A-Z/, /0-9/]},
/gh/
],
/()+$/
).source
);

Personally, I'd go for a less complicated regex:
/\S+#\S+\.\S+/
Sure, it is less accurate than your current pattern, but what are you trying to accomplish? Are you trying to catch accidental errors your users might enter, or are you worried that your users might try to enter invalid addresses? If it's the first, I'd go for an easier pattern. If it's the latter, some verification by responding to an e-mail sent to that address might be a better option.
However, if you want to use your current pattern, it would be (IMO) easier to read (and maintain!) by building it from smaller sub-patterns, like this:
var box1 = "([^<>()[\]\\\\.,;:\s#\"]+(\\.[^<>()[\\]\\\\.,;:\s#\"]+)*)";
var box2 = "(\".+\")";
var host1 = "(\\[[0-9]{1,3}\\.[0-9]{1,3}\\.[0-9]{1,3}\\.[0-9]{1,3}\\])";
var host2 = "(([a-zA-Z\-0-9]+\\.)+[a-zA-Z]{2,})";
var regex = new RegExp("^(" + box1 + "|" + box2 + ")#(" + host1 + "|" + host2 + ")$");

#Hashbrown's great answer got me on the right track. Here's my version, also inspired by this blog.
function regexp(...args) {
function cleanup(string) {
// remove whitespace, single and multi-line comments
return string.replace(/\s+|\/\/.*|\/\*[\s\S]*?\*\//g, '');
}
function escape(string) {
// escape regular expression
return string.replace(/[-.*+?^${}()|[\]\\]/g, '\\$&');
}
function create(flags, strings, ...values) {
let pattern = '';
for (let i = 0; i < values.length; ++i) {
pattern += cleanup(strings.raw[i]); // strings are cleaned up
pattern += escape(values[i]); // values are escaped
}
pattern += cleanup(strings.raw[values.length]);
return RegExp(pattern, flags);
}
if (Array.isArray(args[0])) {
// used as a template tag (no flags)
return create('', ...args);
}
// used as a function (with flags)
return create.bind(void 0, args[0]);
}
Use it like this:
regexp('i')`
//so this is a regex
//here I am matching some numbers
(\d+)
//Oh! See how I didn't need to double backslash that \d?
([a-z]{1,3}) /*note to self, this is group #2*/
`
To create this RegExp object:
/(\d+)([a-z]{1,3})/i

Related

How to split a string by one delimiter but having a particular format as described below

I have a string as:
const str = 'My [Link format](https://google.com) demo'
I want the word array to be like:
['My', '[Link format](https://google.com)', 'demo']
What to do in javascript?
I was trying using split() and str.match(). Nothing worked yet.
This is a simple split on a space as a delimiter, but we us a negative lookahead to check for the combination of open and closed square brackets [] and round brackets ()
const str = 'My [Link format](https://google.com) demo'
console.log(str.split(/\s+(?![^\[]*\])(?![^\(]*\))/));
We also allow for spaces in the URL portion, even though it has a low chance of having spaces, it could still happen
Try it here: https://jsfiddle.net/m4q6e9x7/
["My", "[Link format](https://google.com)", "demo"]
In the fiddle I've tried to show to two separate negative lookaheads for the combination of the types of brackets: (I've put a space in the round brackets to prove the concept)
const str = 'My [Link format](http s://google.com) demo'
ignore space between []
console.log(str.split(/\s+(?![^\[]*\])/));
["My", "[Link format](http", "s://google.com)", "demo"]
ignore space between ()
console.log(str.split(/\s+(?![^\(]*\))/));
["My", "[Link", "format](http s://google.com)", "demo"]
So we can easily combine the two criteria because we need both of them to not match.
Because [] and () need to be escaped, it might be easier to see the regex if we modify and test for spaces between braces {}
const str = 'My {Link format}(https://google.com) demo'
console.log(str.split(/\s+(?![^{]*})/));
["My", "{Link format}(https://google.com)", "demo"]
Both solutions assume, that the string has correct form (meaning basically no space between ']' and '(', no ']' characters inside [...] and similar intuitions. You didn't really provide information about what the input string can be other than your concrete example – so solutions work well in this and very similar cases. Second is very easily modified as needed, first is easily extended to check if the string is in fact not correct.
Solution using Regular Expressions
Below code finds everything before first '[', everything in '[...](...)' pattern (note: first ... must not contain ']', and second – ')', but I assume this would make for an incorrect input in the first place), and everything after that.
So
let regex = /(.*)(\[.*\]\(.*\))(.*)/
let res = str.match(regex).splice(1,3)
gives res as
['My ', '[Link format](https://google.com)', ' demo']
From there, you can trim every entry in this array ('My ' => 'My') for example using a trim function like so:
res.map((val) => val.trim());
Look here for explanation of what the array obtained from .match() method represents, but generally except index 0 it contains capture groups, meaning the parts of string corresponding to parts of regex surrounded by parentheses.
If you are not familiar with Regular Expressions (regexes) in JS, or at all, you will find many online resources about the topic easily. After grasping the basics, regex101 is a nice tool to experiment with regexes and explore their capabilities. When using it, you should probably choose EcmaSCRIPT/JS flavor from the menu on the left.
Equivalent solution without regex
Equivalent solution is to find where is the first '[' manually, as well as where the '[...](...)' pattern ends. Than splice the parts (before '[', pattern, and after pattern) from the string, and probably trim them. So just loop over characters of the string in search of '[' and than ']', '(', ')'. Note that in this case you can easily and granularily decide what to do if the string has unexpected/incorrect form.
TODO: I will probably sketch some code when I have time for it
Regex is your friend!
const regexMdLinks = /!?\[([^\]]*)\]\(([^\)]+)\)/gm
// Example md file contents
const str = `My [Link format](https://google.com) demo My [Link format2](https://google.com/2) demo2`
let regex_splitted = str.split(regexMdLinks);
let arr = [];
//1. Item will be the text (or empty text)
//2. Item is the link text
//3. Item is the url
for(let i = 0; i < regex_splitted.length; i++){
if(i % 3 == 0){ //Split normal text
arr.push(...regex_splitted[i].split(" ").filter(i => i));
} else if(i % 3 == 1){//Add brackets around link text
arr.push("["+regex_splitted[i]+"]");
} else {
arr.push("("+regex_splitted[i]+")");
}
}
console.log(arr)

How to use a variable within a regex in Javascript?

I have a regular expression and would like to put a variable inside it. How do I do?
My code is this:
public regexVariable(vRegex: string, testSentences: Array<any> ) {
const regex = new RegExp('/^.*\b(' + vRegex + ')\b.*$/');
const filterSentece = testSentences.filter(result => {
if (regex.test(result)) {
return result
})
}
const regex = new RegExp(`^.*\\b(${vRegex})\\b.*$`);
You can use template literals (`, instead of "/') to build strings that you can interpolate expresions into; no more oldschool +ing.
The only thing that was an actual issue with your code, though, was the \b character class. This sequence is what you want RegExp to see, but you can't just write that, otherwise you're sending RegExp the backspace character.
You need to write \\b, which as you can see from that link, will make a string with a \ and an ordinary b for RegExp to interpret.
You're almost there, just look at RegEx constructor
const regex = new RegExp('^.*\\b(' + vRegex + ')\\b.*$');

Match regex using array

I made this in C++ and I wanted to convert to JavaScript:
foreach (QString pattern, extensions) {
regex.setPattern(QString("\\.%1").arg(pattern));
regex.setPatternOptions(QRegularExpression::CaseInsensitiveOption);
QRegularExpressionMatch match = regex.match(filename);
if (! match.hasMatch()) continue;
return pattern;
}
It means that foreach extensions (that is an array of extensions) as pattern create a pattern with that to be like: \\.png (for example).
If there's a match it will return the found extension.
I tried to create exactly how I did in C++ but I don't know how to concatenate the returned string from the array to match
const filename = 'example.wutt'
const extensions = ['wutt', 'xnss']
extensions.forEach(pattern => {
const match = filename.match(`\\.${pattern}`)
console.log(match)
})
It does work but it's not case-insensitive as I can't put the i flag.
How can I do that (and if there's a solution using ES6)?
Have a look at How do you use a variable in a regular expression? for building the regex.
If you want to find the extension that matches, you can use Array#find:
const matchedExtension = extensions.find(
ext => new RegExp(String.raw`\.${ext}$`, 'i').test(filename)
);
var extensions = ['png', 'jpeg'];
var filename = 'foo.png';
console.log(extensions.find(
ext => new RegExp(String.raw `\.${ext}$`, 'i').test(filename)
));
Couple of notes:
String.raw is necessary to not treat \. as a string escape sequence but to pass it "as is" to the regular expression engine (alternative you could escape the \, but String.raw is cool).
$ at the end of the pattern ensures that the pattern is only matched at the end of the file name.
If you just want to know whether a pattern matches or not, RegExp#test is the preferred method.
If you are doing this a lot it makes sense to generate an array of regular expressions first (instead of creating the regex every time you call the function).
You can use RegExp constructor with "i" passed as second argument
extensions.forEach(pattern => {
const match = filename.match(new RegExp(`\\.${pattern}$`, "i"));
console.log(match);
})

Doubts in JavaScript RegExp and String.replace() method

I am trying to enter 'username' in a webpage using VBA. So in the source code of the webpage, there are some modifications done to the 'username' value.
I have attached the code,
function myFunction()
{
document.str.value = "Abc02023";
document.str.value = document.str.value.toUpperCase();
pattern = new RegExp("\\*", "g");
document.str.value = document.str.value.replace(pattern, "");
document.str.value = document.str.value.replace(/^\s+/, "");
document.str.value = document.str.value.replace(/\s+$/, "");
}
I read about these and from my understanding, after the modifications document.str.value is ABC02023.
Obviously I am wrong as there would not be no point in doing all these modifications then. Also, I am getting an 'incorrect username error'.
So can anybody please help me to understand these. What would be the value of document.str.value and how did you figure it out? I am new to JavaScript so please forgive me if I am being too slow...
Looks like you are using some very old code to learn from. ☹
Let's see if we can still learn something by bringing this code up to date, then you go find some newer learning materials. Here is a well-written book series with free online versions available: You Don't Know JS.
function myFunction() {
// Assuming your code runs in a browser, `document` is equal to the
// global object. So if in a browser and somewhere outside the function
// a variable `str` has been created, this will add an attribute `value`
// to `str` and set the value of `str.value` to 'Abc02023'. If there is
// no already existing object `document` (for instance not running in
// a browser) or if document does not have an already created property
// called`str` then this will throw a TypeError because you cannot add
// a property to `undefined`.
document.str.value = "Abc02023";
// You probably were just trying to create a new variable `str` so let's
// just start over
}
Second try
function myFunction() {
// create a variable `str` and set it to 'Abc02023'
var str = "Abc02023";
// Take value of str and convert it to all capital letters
// then overwrite current value of str with the result.
// So now `str === 'ABC02023'
str = str.toUpperCase();
// Create a regular expression representing all occurences of `*`
// and assign it to variable `pattern`.
var pattern = new RegExp("\\*", "g");
// Remove all instances of our pattern from the string. (which does not
// affect this string, but prevents user inputting some types of bad
// strings to hack your website.
str = str.replace(pattern, "");
// Remove any leading whitespace form our string (which does not
// affect this string, but cleans up strings input by a user).
str = str.replace(/^\s+/, "");
// Remove any trailing whitespace form our string (which does not
// affect this string, but cleans up strings input by a user).
str = str.replace(/\s+$/, "");
// Let's at least see our result behind the scenes. Press F12
// to see the developer console in most browsers.
console.log("`str` is equal to: ", str );
}
Third try, let's clean this up a little:
// The reason to use functions is so we can contain the logic
// separate from the data. Let's pull extract our data (the string value)
// and then pass it in as a function parameter
var result = myFunction('Abc02023')
console.log('result = ', result)
function myFunction(str) {
str = str.toUpperCase();
// Nicer syntax for defining regular expression.
var pattern = /\*/g;
str = str.replace(pattern, '');
// Unnecesarry use of regular expressions. Let's use trim instead
// to clean leading and trailing whitespace at once.
str = str.trim()
// let's return our result so the rest of the program can use it
// return str
}
Last go round. We can make this much shorter and easier to read by chaining together all the modifications to str. And let's also give our function a useful name and try it out against a bad string.
var cleanString1 = toCleanedCaps('Abc02023')
var cleanString2 = toCleanedCaps(' ** test * ')
console.log('cleanString1 = ', cleanString1)
console.log('cleanString2 = ', cleanString2)
function toCleanedCaps(str) {
return str
.toUpperCase()
.replace(/\\*/g, '')
.trim()
}
#skylize answer is close
what is equivalent to your code is actually
function toCleanedCaps(str) {
return str
.toUpperCase()
.replace(/\*/g, '') // he got this wrong
.trim()
}
Let's go over the statements one by one
document.str.value = document.str.value.toUpperCase();
makes the string uppercase
pattern = new RegExp("\\*", "g");
document.str.value = document.str.value.replace(pattern, "");
replaces between zero and unlimited occurences of the \ character , so no match in this case.
document.str.value = document.str.value.replace(/^\s+/, "");
replaces any whitespace character occurring between one and unlimited times at the beginning of the string, so no match.
document.str.value = document.str.value.replace(/\s+$/, "");
replaces any whitespace character occurring between one and unlimited times at the end of the string, so no match.
You are right. With "Abc02023" as input, the output is what you suggest.

Javascript Regexp Duplicate Line Matching not working correctly

I am writing a Javascript code to parse some grammar files, it is quite some code but I will post relevant information here. I am using Javascript Regexp in order to match a duplicate line held within a string. The string contains, for example (assume the string name is lines):
if
else
;
print
{
}
test1
test1
=
+
-
*
/
(
)
num
string
comment
id
test2
test2
What should happen, is a match found on 'test1' and 'test2'. It should then delete the duplicate, leaving 1 instance of test1 and test2. What is happening is no match at all. I am confident in my regex but javascript may be doing something I am not expecting. Here is the code doing the work on the string given above:
var rex = new RegExp("(.*)(\r?\n\1)+","g");
var re = '/(.*)(\r?\n\1)+/g';
rex.lastIndex = 0;
var m = rex.exec(lines);
if (m) {
alert("Found Duplicate");
var linenum = lines.search(re); //Get line number of error
alert("Error: Symbol Defined twice\n");
alert("Error occured on line: " + linenum);
lines = lines.replace(rex,""); //Gets rid of the duplicate
}
It never gets into the if(m) statement. Therefore no match is found. I tested the regex here: http://regexpal.com/ using the regex in my code as well as the example text provided. It matches just fine, so I am at kind of a loss. If anyone can help, it would be great.
Thank you.
Edit:
Forgot to add, I am testing this in firefox, and it only has to work in firefox. Not sure if that matters.
First error: \ in a JS string is also an escape character.
var rex = new RegExp("(.*)(\r?\n\1)+","g");
should be written
var rex = new RegExp("(.*)(\\r?\\n\\1)+","g");
// or, shorter:
var rex = /(.*)(\r?\n\1)+/g;
if you want to make it work. In the case of the RegExp constructor, you’re passing the pattern as a string to the constructor function. This means you need to escape each \ backslash that occurs in the pattern. If you use a regexp literal, you don’t need to escape them, since they’re not in a string, but retain their ‘normal’ properties in the regexp pattern.
Second error, your expression
var re = '/(.*)(\r?\n\1)+/g';
is wrong. What you’re doing here is assigning a string literal to a variable. I’m assuming you meant to assign a regular expression literal, which should be written like this:
var re = /(.*)(\r?\n\1)+/g;
Third error: the last line
lines = lines.replace(rex,""); //Gets rid of the duplicate
removes both instances of all duplicate lines! If you want to keep the first instance of each duplicate, you should use
lines = lines.replace(rex, "$1");
And finally, this method only detects two consecutive identical lines. Is that what you want, or do you need to detect any duplicates, wherever they may be?
var str = 'if\nelse\n;\nprint\n{\n}\ntest1\ntest1\n=\n+\n-\n*\n/\n(\n)\nnum\nstring\ncomment\nid\ntest2\ntest2\ntest2\ntest2\ntest2';
console.log(str);
str = str.replace(/\r\n?/g,'');
// I prefer replacing all the newline characters with \n's here
str = str.replace(/(^|\n)([^\n]*)(\n\2)+/g,function(m0,m1,m2,m3,ind) {
var line = str.substr(0,ind).split(/\n/).length + 1;
var msg = '[Found duplicate]';
msg += '\nFollowing symbol defined more than once';
msg += '\n\tsymbol: ' + m2;
msg += '\n\ton line ' + line;
console.log(msg);
return m1 + m2;
});
console.log(str);
Otherwise you can skip the first line and change the pattern into
/(^|\r\n?|\n)([^\r\n]*)((?:\r\n?|\n)\2)+/g
Note that [^\n]* will also catch multiple empty lines. If you want to make sure it matches (and replaces) non-empty lines then you might want to use [^\n]+.
[EDIT]
For the record, each m represents each arguments object, so m0 is the whole match, m1 is the 1st subgroup ((^|\n)), m2 is the 2nd subgroup (([^\n]*)) and m3 is the last subgroup ((\n\2)). I could have used arguments[n] instead but these are shorter.
As with the return value, due to lack of lookbehind in the regex flavor used by Javascript, this pattern is catching a possible preceding newline (unless it is the first line) so it needs to return the match and that preceding newline if any. That's why it shouldn't be returning m2 only.

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