CODEWITHSUNDEEP
javascriptregex
JC Grubbs
JC Grubbs

Reputation: 40311

How do you use a variable in a regular expression?

I would like to create a String.replaceAll() method in JavaScript and think using a regex would be the most terse way to do it. However, I can't figure out how to pass a variable into 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 into my regex string?

Upvotes: 1964

Views: 1327691

Answers (29)

trincot
trincot

Reputation: 349908

ECMAScript 2021 introduced replaceAll.

Previously, to replace all occurrences of a plain string in a target string, it was common to pass a regular expression (with g modifier) as argument to the replace method, as passing the string as-is would only replace the first occurrence of that string.

Since ECMAScript 2021, you don't need a regular expression anymore if the goal is to replace all occurrences of a plain string in a target string.

Upvotes: 0

Eric Wendelin
Eric Wendelin

Reputation: 44349

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

// variable == 'REGEX'
let re = new RegExp(String.raw`\s${variable}\s`, "g");

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

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

For older browser or node

// variable == 'REGEX'
var re = new RegExp("\\s" + variable + "\\s", "g");
"mystring1".replace(re, "newstring");

Upvotes: 2429

Maxim Secret
Maxim Secret

Reputation: 31

Easy method:

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

Upvotes: 2

Adam Grant
Adam Grant

Reputation: 13105

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

Upvotes: 1

JBallin
JBallin

Reputation: 9787

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"));

Upvotes: 150

kaptcha
kaptcha

Reputation: 78

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.

Upvotes: -1

Игор Ташевски
Игор Ташевски

Reputation: 671

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.

Upvotes: 3

LeulAria
LeulAria

Reputation: 605

example: regex start with

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

Upvotes: 2

Ajit Hogade
Ajit Hogade

Reputation: 1097

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.

Upvotes: 2

Salman Arshad
Salman Arshad

Reputation: 272006

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]"

Upvotes: 24

keen
keen

Reputation: 850

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!"

Upvotes: 5

Paul Chris Jones
Paul Chris Jones

Reputation: 2909

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>

Upvotes: 3

Victoria Stuart
Victoria Stuart

Reputation: 5072

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.

Upvotes: 0

Fareed Alnamrouti
Fareed Alnamrouti

Reputation: 32154

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>");

Upvotes: 4

Zombo
Zombo

Reputation: 1

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

Upvotes: 40

Ste
Ste

Reputation: 2293

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 :) :) :).

Upvotes: 0

user7396942
user7396942

Reputation:

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 + '})?'
      )

Upvotes: 7

user8094098
user8094098

Reputation:

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'

Upvotes: 3

John Shearing
John Shearing

Reputation: 259

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

Upvotes: 0

bobince
bobince

Reputation: 536319

"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.

Upvotes: 151

MetalGodwin
MetalGodwin

Reputation: 3860

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("]", ">"))

Upvotes: 7

Ry-
Ry-

Reputation: 224862

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.

Upvotes: 2

Alex Li
Alex Li

Reputation: 51

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.

Upvotes: 5

scripto
scripto

Reputation: 2297

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;
    };

Upvotes: 4

Gracenotes
Gracenotes

Reputation: 3042

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".

Upvotes: 285

tvanfosson
tvanfosson

Reputation: 532445

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

Upvotes: 37

unigogo
unigogo

Reputation: 537

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

Upvotes: 10

Jason S
Jason S

Reputation: 189626

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)

Upvotes: 3

Paige Ruten
Paige Ruten

Reputation: 176645

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.

Upvotes: 44

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