How do you use a variable in a regular expression?

Asked 2023-09-21 08:08:30 View 139,318

I would like to create a String.replaceAll() method in JavaScript and I'm thinking that using a regex would be most terse way to do it. However, I can't figure out how to pass a variable in to a regex. I can do this already which will replace all the instances of "B" with "A".

"ABABAB".replace(/B/g, "A");

But I want to do something like this:

String.prototype.replaceAll = function(replaceThis, withThis) {
    this.replace(/replaceThis/g, withThis);
};

But obviously this will only replace the text "replaceThis"...so how do I pass this variable in to my regex string?

Answers

Instead of using the /regex\d/g syntax, you can construct a new RegExp object:

var replace = "regex\\d";
var re = new RegExp(replace,"g");

You can dynamically create regex objects this way. Then you will do:

"mystring1".replace(re, "newstring");

Answered   2023-09-21 08:08:30

  • If you need to use an expression like /\/word\:\w*$/, be sure to escape your backslashes: new RegExp( '\\/word\\:\\w*$' ). - anyone
  • The question suggests that the RegEx is only used to do a constant string replacement. So this is answer is wrong as it would fail if the string contains RegEx meta characters. Sad it is voted this high, will make many headaches... - anyone
  • An example of this passing a variable would make this a good answer. I'm still struggling after reading this. - anyone

As Eric Wendelin mentioned, you can do something like this:

str1 = "pattern"
var re = new RegExp(str1, "g");
"pattern matching .".replace(re, "regex");

This yields "regex matching .". However, it will fail if str1 is ".". You'd expect the result to be "pattern matching regex", replacing the period with "regex", but it'll turn out to be...

regexregexregexregexregexregexregexregexregexregexregexregexregexregexregexregexregexregex

This is because, although "." is a String, in the RegExp constructor it's still interpreted as a regular expression, meaning any non-line-break character, meaning every character in the string. For this purpose, the following function may be useful:

 RegExp.quote = function(str) {
     return str.replace(/([.?*+^$[\]\\(){}|-])/g, "\\$1");
 };

Then you can do:

str1 = "."
var re = new RegExp(RegExp.quote(str1), "g");
"pattern matching .".replace(re, "regex");

yielding "pattern matching regex".

Answered   2023-09-21 08:08:30

  • You know that the first parameter to replace can be a normal string and don't have to be a regexp? str1 = "."; alert("pattern matching .".replace(str1, "string")); - anyone
  • @some: of course. That's because the above example is trivial. When you need to search for or replace a pattern combined with a regular string, do str.match(new RegExp("https?://" + RegExp.escape(myDomainName)), for instance. It's annoying that the escape function is not built in. - anyone
  • developer.mozilla.org/en-US/docs/JavaScript/Guide/… offers a similar function, but they exclude -, and include =!:/. - anyone
  • The correct term is "escape", not "quote". Just BTW. - anyone
  • Would /[^\w\s]/g be a safe alternative to /([.?*+^$[\]\\(){}|-])/g? - anyone

"ABABAB".replace(/B/g, "A");

As always: don't use regex unless you have to. For a simple string replace, the idiom is:

'ABABAB'.split('B').join('A')

Then you don't have to worry about the quoting issues mentioned in Gracenotes's answer.

Answered   2023-09-21 08:08:30

  • And have you measured that this is faster than regex? - anyone
  • This seems preferable, especially when needing to match on special regex characters like '.' - anyone
  • Uhm... Doesn't split take a RegExp too; if so, wouldn't it cause the same problem ? Anyway... .split().join() may be slower on some platforms, because it's two operations, whereas .replace() is one operation and may be optimized. - anyone
  • @PacMan--: both split and replace can take either a string or a RegExp object. The problem that replace has that split doesn't is that when you use a string you only get a single replacement. - anyone
  • benchmark here: jsperf.com/replace-vs-split-join-vs-replaceall/23 - anyone

If you want to get all occurrences (g), be case insensitive (i), and use boundaries so that it isn't a word within another word (\\b):

re = new RegExp(`\\b${replaceThis}\\b`, 'gi');

let inputString = "I'm John, or johnny, but I prefer john.";
let replaceThis = "John";
let re = new RegExp(`\\b${replaceThis}\\b`, 'gi');
console.log(inputString.replace(re, "Jack"));

Answered   2023-09-21 08:08:30

  • thank you! (afaict, yours is the only answer explicitly with Emacs/rx-style interpolation, via template strings.) - anyone
  • What about replaceAll? Would it work the same as replace with the global flag? - anyone
  • @cezar technically you could use replaceAll with the exact regex above (including global flag) - but it would have no benefit. You would get an error if you tried to use it without the global flag, see this. - anyone
  • hi i'm trying to use this but not working 'const regex = new RegExp(/(?=.{\\b${digits}\\b}).*/g);' whereas digits is a numeric variable I'm passing down as a parameter. If possible can u explain how can I fix this? - anyone
  • @joekevinrayan96 integers work fine when I test them with my current example. Please create a separate question with a minimal reproducible example. - anyone

This:

var txt=new RegExp(pattern,attributes);

is equivalent to this:

var txt=/pattern/attributes;

See http://www.w3schools.com/jsref/jsref_obj_regexp.asp.

Answered   2023-09-21 08:08:30

  • yep, but in first example it uses pattern as variable, in 2nd as a string - anyone

For anyone looking to use a variable with the match method, this worked for me:

var alpha = 'fig';
'food fight'.match(alpha + 'ht')[0]; // fight

Answered   2023-09-21 08:08:30

this.replace( new RegExp( replaceThis, 'g' ), withThis );

Answered   2023-09-21 08:08:30

  • I like this answer as it doesn't create the extra (& pointless) variable. - anyone

You need to build the regular expression dynamically and for this you must use the new RegExp(string) constructor with escaping.

There is a built-in function in jQuery UI autocomplete widget called $.ui.autocomplete.escapeRegex:

It'll take a single string argument and escape all regex characters, making the result safe to pass to new RegExp().

If you are not using jQuery UI you can copy its definition from the source:

function escapeRegex( value ) {
    return value.replace( /[\-\[\]{}()*+?.,\\\^$|#\s]/g, "\\$&" );
}

And use it like this:

"[z-a][z-a][z-a]".replace(new RegExp(escapeRegex("[z-a]"), "g"), "[a-z]");
//            escapeRegex("[z-a]")       -> "\[z\-a\]"
// new RegExp(escapeRegex("[z-a]"), "g") -> /\[z\-a\]/g
// end result                            -> "[a-z][a-z][a-z]"

Answered   2023-09-21 08:08:30

String.prototype.replaceAll = function (replaceThis, withThis) {
   var re = new RegExp(replaceThis,"g"); 
   return this.replace(re, withThis);
};
var aa = "abab54..aba".replaceAll("\\.", "v");

Test with this tool

Answered   2023-09-21 08:08:30

You can use a string as a regular expression. Don’t forget to use new RegExp.

Example:

var yourFunction = new RegExp(
        '^-?\\d+(?:\\.\\d{0,' + yourVar + '})?'
      )

Answered   2023-09-21 08:08:30

String.prototype.replaceAll = function(a, b) {
    return this.replace(new RegExp(a.replace(/([.?*+^$[\]\\(){}|-])/ig, "\\$1"), 'ig'), b)
}

Test it like:

var whatever = 'Some [b]random[/b] text in a [b]sentence.[/b]'

console.log(whatever.replaceAll("[", "<").replaceAll("]", ">"))

Answered   2023-09-21 08:08:30

To satisfy my need to insert a variable/alias/function into a Regular Expression, this is what I came up with:

oldre = /xx\(""\)/;
function newre(e){
    return RegExp(e.toString().replace(/\//g,"").replace(/xx/g, yy), "g")
};

String.prototype.replaceAll = this.replace(newre(oldre), "withThis");

where 'oldre' is the original regexp that I want to insert a variable, 'xx' is the placeholder for that variable/alias/function, and 'yy' is the actual variable name, alias, or function.

Answered   2023-09-21 08:08:30

  • After trying every single solution for inserting a variable inside the regular expression, yours was the only one that worked for me. Thank you sooo much! - anyone

And the CoffeeScript version of Steven Penny's answer, since this is #2 Google result....even if CoffeeScript is just JavaScript with a lot of characters removed...;)

baz = "foo"
filter = new RegExp(baz + "d")
"food fight".match(filter)[0] // food

And in my particular case:

robot.name = hubot
filter = new RegExp(robot.name)
if msg.match.input.match(filter)
  console.log "True!"

Answered   2023-09-21 08:08:30

  • why a downvote? coffeescript -IS- javascript with it's own specific syntax. - anyone
  • robot.name=hubot is not javascript. - anyone

Here's another replaceAll implementation:

    String.prototype.replaceAll = function (stringToFind, stringToReplace) {
        if ( stringToFind == stringToReplace) return this;
        var temp = this;
        var index = temp.indexOf(stringToFind);
        while (index != -1) {
            temp = temp.replace(stringToFind, stringToReplace);
            index = temp.indexOf(stringToFind);
        }
        return temp;
    };

Answered   2023-09-21 08:08:30

You can use this if $1 does not work for you:

var pattern = new RegExp("amman", "i");
"abc Amman efg".replace(pattern, "<b>" + "abc Amman efg".match(pattern)[0] + "</b>");

Answered   2023-09-21 08:08:30

While you can make dynamically-created RegExp's (as per the other responses to this question), I'll echo my comment from a similar post: The functional form of String.replace() is extremely useful and in many cases reduces the need for dynamically-created RegExp objects. (which are kind of a pain 'cause you have to express the input to the RegExp constructor as a string rather than use the slashes /[A-Z]+/ regexp literal format)

Answered   2023-09-21 08:08:30

This self calling function will iterate over replacerItems using an index, and change replacerItems[index] globally on the string with each pass.

  const replacerItems = ["a", "b", "c"];    

    function replacer(str, index){
          const item = replacerItems[index];
          const regex = new RegExp(`[${item}]`, "g");
          const newStr = str.replace(regex, "z");
          if (index < replacerItems.length - 1) {
            return replacer(newStr, index + 1);
          }
          return newStr;
    }

// console.log(replacer('abcdefg', 0)) will output 'zzzdefg'

Answered   2023-09-21 08:08:30

None of these answers were clear to me. I eventually found a good explanation at How to use a variable in replace function of JavaScript

The simple answer is:

var search_term = new RegExp(search_term, "g");
text = text.replace(search_term, replace_term);

For example:

$("button").click(function() {
  Find_and_replace("Lorem", "Chocolate");
  Find_and_replace("ipsum", "ice-cream");
});

function Find_and_replace(search_term, replace_term) {
  text = $("textbox").html();
  var search_term = new RegExp(search_term, "g");
  text = text.replace(search_term, replace_term);
  $("textbox").html(text);
}
<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<textbox>
  Lorem ipsum Lorem ipsum Lorem ipsum Lorem ipsum Lorem ipsum Lorem ipsum
</textbox>
<button>Click me</button>

Answered   2023-09-21 08:08:30

  • You're overwriting a closure variable, no need to use var here. Also, if you pass \b or \1 it would break. - anyone

I found so many answers with weird examples in here and in other open tickets on stackoverflow or similar forums.

This is the simplest option in my opinion how u can put variable as template literal string;

const someString = "abc";
const regex = new RegExp(`^ someregex ${someString} someregex $`);

As u can see I'm not puting forward slash at the beginning or the end, the RegExp constructor will reconstruct the valid regex literal. Works with yup matches function also.

Answered   2023-09-21 08:08:30

  • youre not using slash, but youre using a backtick? Maybe you can explain this. - anyone

You can always use indexOf repeatedly:

String.prototype.replaceAll = function(substring, replacement) {
    var result = '';
    var lastIndex = 0;

    while(true) {
        var index = this.indexOf(substring, lastIndex);
        if(index === -1) break;
        result += this.substring(lastIndex, index) + replacement;
        lastIndex = index + substring.length;
    }

    return result + this.substring(lastIndex);
};

This doesn’t go into an infinite loop when the replacement contains the match.

Answered   2023-09-21 08:08:30

One way to implement is by taking the value from a text field which is the one you want to replace and another is the "replace with" text field, getting the value from text-field in a variable and setting the variable to RegExp function to further replace. In my case I am using jQuery, but you can also do it by only JavaScript too.

JavaScript code:

  var replace =document.getElementById("replace}"); // getting a value from a text field with I want to replace
  var replace_with = document.getElementById("with"); //Getting the value from another text fields with which I want to replace another string.

  var sRegExInput = new RegExp(replace, "g");
  $("body").children().each(function() {
    $(this).html($(this).html().replace(sRegExInput,replace_with));
  });

This code is on the Onclick event of a button, and you can put this in a function to call.

So now you can pass a variable in the replace function.

Answered   2023-09-21 08:08:30

  • Your replace_with variable will contain the DOM element not the value itself - anyone
  • The link is broken: "Page not found - Mindfire Solutions. 404. Looks like you are lost." - anyone
  • Just ignor the link you can use the code given in the answer it should work accordingly - anyone

example: regex start with

function startWith(char, value) {
    return new RegExp(`^[${char}]`, 'gi').test(value);
}

Answered   2023-09-21 08:08:30

For multiple replace without regular expressions I went with the following:

      let str = "I am a cat man. I like cats";
      let find = "cat";
      let replace = "dog";


      // Count how many occurrences there are of the string to find 
      // inside the str to be examined.
      let findCount = str.split(find).length - 1;

      let loopCount = 0;

      while (loopCount < findCount) 
      {
        str = str.replace(find, replace);
        loopCount = loopCount + 1;
      }  

      console.log(str);
      // I am a dog man. I like dogs

The important part of the solution was found here

Answered   2023-09-21 08:08:30

If you pass the variable with the correct syntax, you can do this like so with the code below.

This has the added benefit of using the flags in the same variable.

Also you don't have to double escape \ in the regular expression when it comes to \w, etc.

var str = 'regexVariable example: This is my example of RegExp replacing with a regexVariable.'
var reVar = /(.*?)(regex\w+?iable)(.+?)/gi;
var resStr = str.replace(new RegExp(reVar), '$1 :) :) :) $2 :) :) :)$3');
console.log(resStr);

// Returns:
// :) :) :) regexVariable :) :) :) example: This is my example of RegExp replacing with a  :) :) :) regexVariable :) :) :).

The prototype version as per the OP's example:

var str = 'regexVariable prototype: This is my example of RegExp replacing with a regexVariable.'

String.prototype.regexVariable = function(reFind, reReplace) {
return str.replace(new RegExp(reFind), reReplace);
}

var reVar = /(.*?)(regex\w+?iable)(.+?)/gi;

console.log(str.regexVariable(reVar, '$1 :) :) :) $2 :) :) :)$3'));

// Returns:
// :) :) :) regexVariable :) :) :) prototype: This is my example of replacing with a  :) :) :) regexVariable :) :) :).

Answered   2023-09-21 08:08:30

As a relative JavaScript novice, the accepted answer https://stackoverflow.com/a/494046/1904943 is noted / appreciated, but it is not very intuitive.

Here is a simpler interpretation, by example (using a simple JavaScript IDE).

myString = 'apple pie, banana loaf';

console.log(myString.replaceAll(/pie/gi, 'PIE'))
// apple PIE, banana loaf

console.log(myString.replaceAll(/\bpie\b/gi, 'PIE'))
// apple PIE, banana loaf

console.log(myString.replaceAll(/pi/gi, 'PIE'))
// apple PIEe, banana loaf

console.log(myString.replaceAll(/\bpi\b/gi, 'PIE'))
// [NO EFFECT] apple pie, banana loaf

const match_word = 'pie';

console.log(myString.replaceAll(/match_word/gi, '**PIE**'))
// [NO EFFECT] apple pie, banana loaf

console.log(myString.replaceAll(/\b`${bmatch_word}`\b/gi, '**PIE**'))
// [NO EFFECT] apple pie, banana loaf

// ----------------------------------------
// ... new RegExp(): be sure to \-escape your backslashes: \b >> \\b ...

const match_term = 'pie';
const match_re = new RegExp(`(\\b${match_term}\\b)`, 'gi')

console.log(myString.replaceAll(match_re, 'PiE'))
// apple PiE, banana loaf

console.log(myString.replace(match_re, '**PIE**'))
// apple **PIE**, banana loaf

console.log(myString.replaceAll(match_re, '**PIE**'))
// apple **PIE**, banana loaf

Application

E.g.: replacing (color highlighting) words in string / sentence, [optionally] if the search term matches a more than a user-defined proportion of the matched word.

Note: original character case of matched term is retained. hl: highlight; re: regex | regular expression

mySentence = "Apple, boOk? BOoks; booKEd. BookMark, 'BookmarkeD', bOOkmarks! bookmakinG, Banana; bE, BeEn, beFore."

function replacer(mySentence, hl_term, hl_re) {
    console.log('mySentence [raw]:', mySentence)
    console.log('hl_term:', hl_term, '| hl_term.length:', hl_term.length)
    cutoff = hl_term.length;
    console.log('cutoff:', cutoff)

    // `.match()` conveniently collects multiple matched items
    // (including partial matches) into an [array]
    const hl_terms  = mySentence.toLowerCase().match(hl_re, hl_term);
    if (hl_terms == null) {
        console.log('No matches to hl_term "' + hl_term + '"; echoing input string then exiting ...')
        return mySentence;
    }
    console.log('hl_terms:', hl_terms)
    for (let i = 0;  i < hl_terms.length; i++) {
        console.log('----------------------------------------')
        console.log('[' + i + ']:', hl_terms[i], '| length:', hl_terms[i].length, '| parseInt(0.7(length)):', parseInt(0.7*hl_terms[i].length))
        // TEST: if (hl_terms[i].length >= cutoff*10) {
        if (cutoff >= parseInt(0.7 * hl_terms[i].length)) {
            var match_term = hl_terms[i].toString();

            console.log('matched term:', match_term, '[cutoff length:', cutoff, '| 0.7(matched term length):', parseInt(0.7 * hl_terms[i].length))

            const match_re = new RegExp(`(\\b${match_term}\\b)`, 'gi')

            mySentence = mySentence.replaceAll(match_re, '<font style="background:#ffe74e">$1</font>');
        }
        else {
            var match_term = hl_terms[i].toString();
            console.log('NO match:', match_term, '[cutoff length:', cutoff, '| 0.7(matched term length):', parseInt(0.7 * hl_terms[i].length))
        }
    }
    return mySentence;
}

// TESTS:
// const hl_term = 'be';
// const hl_term = 'bee';
// const hl_term = 'before';
// const hl_term = 'book';
const hl_term = 'bookma';
// const hl_term = 'Leibniz';

// This regex matches from start of word:
const hl_re = new RegExp(`(\\b${hl_term}[A-z]*)\\b`, 'gi')

mySentence = replacer(mySentence, hl_term, hl_re);
console.log('mySentence [processed]:', mySentence)

Output

mySentence [raw]: Apple, boOk? BOoks; booKEd. BookMark, 'BookmarkeD',
bOOkmarks! bookmakinG, Banana; bE, BeEn, beFore.

hl_term: bookma | hl_term.length: 6
cutoff: 6
hl_terms: Array(4) [ "bookmark", "bookmarked", "bookmarks", "bookmaking" ]

----------------------------------------
[0]: bookmark | length: 8 | parseInt(0.7(length)): 5
matched term: bookmark [cutoff length: 6 | 0.7(matched term length): 5
----------------------------------------
[1]: bookmarked | length: 10 | parseInt(0.7(length)): 7
NO match: bookmarked [cutoff length: 6 | 0.7(matched term length): 7
----------------------------------------
[2]: bookmarks | length: 9 | parseInt(0.7(length)): 6
matched term: bookmarks [cutoff length: 6 | 0.7(matched term length): 6
----------------------------------------
[3]: bookmaking | length: 10 | parseInt(0.7(length)): 7
NO match: bookmaking [cutoff length: 6 | 0.7(matched term length): 7

mySentence [processed]: Apple, boOk? BOoks; booKEd.
<font style="background:#ffe74e">BookMark</font>, 'BookmarkeD',
<font style="background:#ffe74e">bOOkmarks</font>! bookmakinG,
Banana; bE, BeEn, beFore.

Answered   2023-09-21 08:08:30

In case anyone else was looking for this, here's how you keep the operators:

// BAD
let foo = "foo"
new RegExp(`${foo}\s`, "g");
// => /foos/g

// GOOD
let foo = "foo"
new RegExp(`${foo}${/\s/.source}`, "g");
// => /foo\s/g

Answered   2023-09-21 08:08:30

All these answers seem extremely complicated, when there is a much simpler answer that still gets the job done using regex.

String.prototype.replaceAll = function(replaceThis, withThis) {
    const expr = `${replaceThis}`
    this.replace(new RegExp(expr, "g"), withThis);
};

Explanation

The RegExp constructor takes 2 arguments: the expression, and flags. By using a template string in the expression, we can pass in the variable into the class, and it will transform it to be /(value of the replaceThis variable)/g.

Answered   2023-09-21 08:08:30