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How to remove a specific character from a string and put it back later when needed

For example
let myString = "This is my string";
let replacedString = myString.replace(/\ /g, "") //Thisismystring
Now that all the whitespaces have been removed, how do I put them back in the exact position?
Additionally, let's suppose the replaced string undergoes some change and becomes
let myChangedString = "(T)(h)(i)(s)(i)(s)(m)(y)(s)(t)(r)(i)(n)(g)";
Now I want to put the whitespaces back where they used to be i.e after (s) and before (i), after (s) and before (m), after (y) before (s)
I've spent a couple of hours on this and been stuck in the same position, any form of help would be greatly appreciated.
EDIT: Solved, thank you very much.

The trick here is to replace the spaces with another character - rather than just removing the space. That way - its a simple matter to replace the added character with a space to return the spaces to where they started. When I do this - I always use the tilde character "~" since it is easily recognisable as well as unlikely to actually be used in a string.
I have added a few variations / modifications as well as the example you have provided with every character being wrapped in parentheses - just note that you will need to escape these when replaceing the (~) for the " " space character.
let myString = "This is my string";
let replacedString = myString.replace(/\ /g, "~");
console.log(replacedString);//This~is~my~string
let modifiedString = replacedString.replace(/my/g, "your");
console.log(modifiedString);//This~is~your~string
let spacedString = modifiedString.replace(/~/g, " ");
console.log(spacedString);//This is your string
// using your example of wrapping each character in parentheses
let myChangedString = "(" + modifiedString.split('').join(")(") + ")";
console.log(myChangedString); //(T)(h)(i)(s)(~)(i)(s)(~)(y)(o)(u)(r)(~)(s)(t)(r)(i)(n)(g)
let mySpacedString = myChangedString.replace(/\(~\)/g, " ");
console.log(mySpacedString); //(T)(h)(i)(s) (i)(s) (y)(o)(u)(r) (s)(t)(r)(i)(n)(g)

Why not replace only the parts you need to be replaced?
For example search for word character and replace with the wanted parts.
console.log("This is my string".replace(/\w/g, '($&)'));

Better you just transform your original array. Loop through array and modify the char is not empty.
let myString = "This is my string";
let chars = [...myString].map(item => item !== ' ' ? '(' + item + ')': item)
console.log(chars.join(''))

Are you looking for this...
var result = "thisismystring".replace(/^(.{4})(.{2})(.{2})(.*)$/, "$1 $2 $3 $4");
alert(result);

regex to remove number (year only) from string

I know the regex that separates two words as following:
input:
'WonderWorld'
output:
'Wonder World'
"WonderWorld".replace(/([A-Z])/g, ' $1');
Now I am looking to remove number in year format from string, what changes should be done in the above code to get:
input
'WonderWorld 2016'
output
'Wonder World'

You can match the location before an uppercase letter (but excluding the beginning of a line) with \B(?=[A-Z]) and match the trailing spaces if any with 4 digits right before the end (\s*\b\d{4}\b). In a callback, check if the match is not empty, and replace accordingly. If a match is empty, we matched the location before an uppercase letter (=> replace with a space) and if not, we matched the year at the end (=> replace with empty string). The four digit chunks are only matched as whole words due to the \b word boundaries around the \d{4}.
var re = /\B(?=[A-Z])|\s*\d{4}\b/g;
var str = 'WonderWorld 2016';
var result = str.replace(re, function(match) {
return match ? "" : " ";
});
document.body.innerHTML = "<pre>'" + result + "'</pre>";
A similar approach, just a different pattern for matching glued words (might turn out more reliable):
var re = /([a-z])(?=[A-Z])|\s*\b\d{4}\b/g;
var str = 'WonderWorld 2016';
var result = str.replace(re, function(match, group1) {
return group1 ? group1 + " " : "";
});
document.body.innerHTML = "<pre>'" + result + "'</pre>";
Here, ([a-z])(?=[A-Z]) matches and captures into Group 1 a lowercase letter that is followed with an uppercase one, and inside the callback, we check if Group 1 matched (with group1 ?). If it matched, we return the group1 + a space. If not, we matched the year at the end, and remove it.

Try this:
"WonderWorld 2016".replace(/([A-Z])|\b[0-9]{4}\b/g, ' $1')

How about this, a single regex to do what you want:
"WonderWorld 2016".replace(/([A-Z][a-z]+)([A-Z].*)\s.*/g, '$1 $2');
"Wonder World"
get everything apart from digits and spaces.

re-code of #Wiktor Stribiżew's solution:
str can be any "WonderWorld 2016" | "OneTwo 1000 ThreeFour" | "Ruby 1999 IamOnline"
str.replace(/([a-z])(?=[A-Z])|\s*\d{4}\b/g, function(m, g) {
return g ? g + " " : "";
});

import re
remove_year_regex = re.compile(r"[0-9]{4}")
Test regex expression here

Javascript Regular Expression to search a multiline textarea for an expression

Let's say there is a textarea with the following value ('*' being used as a bullet point):
*south
*north
*west
I want to be able to automatically generate an array of these words using Regular Expression, like this.
["south","north","west"]
Below is the expression I tried.
/\*.*/gm.exec(text)
Unfortunately it returns this instead.
["*south"]
Apparently, RegExp recognizes there is a line break such that it only returns the first item, yet it doesn't pick up the 2nd and 3rd lines.
/\*.*/gm.exec('*south \n *north')
This also has the same result.

You need to tell the regex engine to match at the beginning of a line with ^, and capture the part after the first * with a pair of unescaped parentheses. Then, you can use RegExp#exec() in a loop, and get the value you need in Group 1. The ^\s*\*\s*(.*) regex matches:
^ - start of a line (due to /m multiline modifier)
\s* - zero or more whitespace symbols
\* - a literal asterisk
\s* - again, optional whitespace(s)
(.*) - zero or more characters other than a newline.
var re = /^\s*\*\s*(.*)/gm;
var str = '*south\n *north\n* west ';
var res = [];
while ((m = re.exec(str)) !== null) {
res.push(m[1]);
}
document.write("<pre>" + JSON.stringify(res, 0, 4) + "</pre>");
Another solution:
Split with newline (a regex is possible here if there can be \r or \n) and then get rid of the initial *:
var str = '*south\n*north\n*west ';
var res = [];
str.split(/[\r\n]+/).forEach(function(e) {
res.push(e.replace(/^\s*\*\s*/, ''));
});
document.write("<pre>" + JSON.stringify(res, 0, 4) + "</pre>");

#VKS solution works, but if it is not mandatory to use regex then try this fiddle
<textarea id="textA1"></textarea>
$( "#textA1" ).blur( function(){
var value = $( this ).val();
console.log( value.split( "\n" ) );
} )

You will have to run a loop.
var re = /\*(.*)/gm;
var str = '*south\n*north\n*west ';
var m;
while ((m = re.exec(str)) !== null) {
// View your result using the m-variable.
// eg m[0] etc.
}
See demo.
https://regex101.com/r/iJ7bT6/11
or you an split by (?=\*).See demo.
https://regex101.com/r/iJ7bT6/12

Remove ALL white spaces from text

$("#topNav" + $("#breadCrumb2nd").text().replace(" ", "")).addClass("current");
This is a snippet from my code. I want to add a class to an ID after getting another ID's text property. The problem with this, is the ID holding the text I need, contains gaps between the letters.
I would like the white spaces removed. I have tried TRIM()and REPLACE() but this only partially works. The REPLACE() only removes the 1st space.

You have to tell replace() to repeat the regex:
.replace(/ /g,'')
The g character makes it a "global" match, meaning it repeats the search through the entire string. Read about this, and other RegEx modifiers available in JavaScript here.
If you want to match all whitespace, and not just the literal space character, use \s instead:
.replace(/\s/g,'')
You can also use .replaceAll if you're using a sufficiently recent version of JavaScript, but there's not really any reason to for your specific use case, since catching all whitespace requires a regex, and when using a regex with .replaceAll, it must be global, so you just end up with extra typing:
.replaceAll(/\s/g,'')

.replace(/\s+/, "")
Will replace the first whitespace only, this includes spaces, tabs and new lines.
To replace all whitespace in the string you need to use global mode
.replace(/\s/g, "")

Now you can use "replaceAll":
console.log(' a b c d e f g '.replaceAll(' ',''));
will print:
abcdefg
But not working in every possible browser:
https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/String/replaceAll

Regex for remove white space
\s+
var str = "Visit Microsoft!";
var res = str.replace(/\s+/g, "");
console.log(res);
or
[ ]+
var str = "Visit Microsoft!";
var res = str.replace(/[ ]+/g, "");
console.log(res);
Remove all white space at begin of string
^[ ]+
var str = " Visit Microsoft!";
var res = str.replace(/^[ ]+/g, "");
console.log(res);
remove all white space at end of string
[ ]+$
var str = "Visit Microsoft! ";
var res = str.replace(/[ ]+$/g, "");
console.log(res);
var mystring="fg gg";
console.log(mystring.replaceAll(' ',''))

** 100% working
use replace(/ +/g,'_'):
let text = "I love you"
text = text.replace( / +/g, '_') // replace with underscore ('_')
console.log(text) // I_love_you

Using String.prototype.replace with regex, as mentioned in the other answers, is certainly the best solution.
But, just for fun, you can also remove all whitespaces from a text by using String.prototype.split and String.prototype.join:
const text = ' a b c d e f g ';
const newText = text.split(/\s/).join('');
console.log(newText); // prints abcdefg

I don't understand why we need to use regex here when we can simply use replaceAll
let result = string.replaceAll(' ', '')
result will store string without spaces

let str = 'a big fat hen clock mouse '
console.log(str.split(' ').join(''))
// abigfathenclockmouse

Use string.replace(/\s/g,'')
This will solve the problem.
Happy Coding !!!

simple solution could be : just replace white space ask key value
val = val.replace(' ', '')

Use replace(/\s+/g,''),
for example:
const stripped = ' My String With A Lot Whitespace '.replace(/\s+/g, '')// 'MyStringWithALotWhitespace'

Well, we can also use that [^A-Za-z] with g flag for removing all the spaces in text. Where negated or complemente or ^. Show to the every character or range of character which is inside the brackets. And the about g is indicating that we search globally.
let str = "D S# D2m4a r k 23";
// We are only allowed the character in that range A-Za-z
str = str.replace(/[^A-Za-z]/g,""); // output:- DSDmark
console.log(str)
javascript - Remove ALL white spaces from text - Stack Overflow

Using .replace(/\s+/g,'') works fine;
Example:
this.slug = removeAccent(this.slug).replace(/\s+/g,'');

function RemoveAllSpaces(ToRemove)
{
let str = new String(ToRemove);
while(str.includes(" "))
{
str = str.replace(" ", "");
}
return str;
}

Replace a substring with javascript

Need to replace a substring in URL (technically just a string) with javascript.
The string like
http://blah-blah.com/search?par_one=test&par_two=anothertest&SearchableText=TO_REPLACE
or
http://blah-blah.com/search?par_one=test&SearchableText=TO_REPLACE&par_two=anothertest
means, the word to replace can be either at the most end of the URL or in the middle of it.
I am trying to cover these with the following:
var newWord = NEW_SEARCH_TERM;
var str = 'http://blah-blah.com/search?par_one=test&SearchableText=TO_REPLACE&par_two=anothertest';
var regex = /^\S+SearchableText=(.*)&?\S*$/;
str = str.replace(regex, newWord);
But no matter what I do I get str = NEW_SEARCH_TERM. Moreover the regular expression when I try it in RegExhibit, selects the word to replace and everything that follows it that is not what I want.
How can I write a universal expression to cover both cases and make the correct string be saved in the variable?

str.replace(/SearchableText=[^&]*/, 'SearchableText=' + newWord)

The \S+ and \S* in your regex match all non-whitespace characters.
You probably want to remove them and the anchors.

http://jsfiddle.net/mplungjan/ZGbsY/
ClyFish did it while I was fiddling
var url1="http://blah-blah.com/search?par_one=test&par_two=anothertest&SearchableText=TO_REPLACE";
var url2 ="http://blah-blah.com/search?par_one=test&SearchableText=TO_REPLACE&par_two=anothertest"
var newWord = "foo";
function replaceSearch(str,newWord) {
var regex = /SearchableText=[^&]*/;
return str.replace(regex, "SearchableText="+newWord);
}
document.write(replaceSearch(url1,newWord))
document.write('<hr>');
document.write(replaceSearch(url2,newWord))