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?
const re = new RegExp(`${replaceThis}`, 'g'); str.replace(re, withThis);
- anyone Instead of using the /regex\d/g
syntax, you can construct a new RegExp object:
var replace = "regex\\d";
var re = new RegExp(replace,"g");
You can dynamically create regex objects this way. Then you will do:
"mystring1".replace(re, "newstring");
Answered 2023-09-21 08:08:30
/\/word\:\w*$/
, be sure to escape your backslashes: new RegExp( '\\/word\\:\\w*$' )
. - anyone As Eric Wendelin mentioned, you can do something like this:
str1 = "pattern"
var re = new RegExp(str1, "g");
"pattern matching .".replace(re, "regex");
This yields "regex matching ."
. However, it will fail if str1 is "."
. You'd expect the result to be "pattern matching regex"
, replacing the period with "regex"
, but it'll turn out to be...
regexregexregexregexregexregexregexregexregexregexregexregexregexregexregexregexregexregex
This is because, although "."
is a String, in the RegExp constructor it's still interpreted as a regular expression, meaning any non-line-break character, meaning every character in the string. For this purpose, the following function may be useful:
RegExp.quote = function(str) {
return str.replace(/([.?*+^$[\]\\(){}|-])/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
-
, and include =!:/
. - anyone /[^\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
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 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
rx
-style interpolation, via template strings.) - anyone replaceAll
? Would it work the same as replace
with the global flag? - anyone 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 \\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 This:
var txt=new RegExp(pattern,attributes);
is equivalent to this:
var txt=/pattern/attributes;
Answered 2023-09-21 08:08:30
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
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
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
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
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
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
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
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