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How do I delete all characters starting from a specific text on a string

The best way to explain this is by example. I'm using jQuery to do this.
Example I have a string
var str = "1.) Ben"
how can I dynamically omit the character 1.) including the space such that str === "Ben"
str can be dynamic such that order can increment from ones, tens, to hundreds.
E.G.
var str = "52.) Ken Bush"
or
var str = "182.) Hailey Quen"
Expected output
str === "Ken Bush"
or
str === "Hailey Quen"

Example
var str = "182.) Hailey Quen"
var test = str.split(') ');
test = test[1];
//output "Hailey Quen"

You can use regex replacement to get what you want.
var str = "182.) Hailey"
var newStr = str.replace(/^\d+\.\)\s*/, '')
// Hailey
var s = "1456.) Hard Spocker".replace(/^\d+\.\)\s*/, '')
// Hard Spocker
^ makes sure that the pattern is matched at the start of the string only
\d+ will match one or more digits.
\. will match the . with escaping
) is a symbol so we need to escape it using \ as \)
\s* will match one or more spaces.
You can learn about these symbols here.

Try using .substring() and .indexOf() as shown :-
var str = "182.) Hailey Quen"
alert(str.substring(str.indexOf(' ')))
DEMO
OR use .split() as shown :-
var str = "182.) Hailey Quen"
alert($.trim(str.split(')')[1]))
DEMO

You can do it regular expression,
var str = "52.) Ken".replace(/\d+\.\)\s/g,"");
console.log(str); //Ken
DEMO
If you have zero or more than zero spaces after the ) symbol then you can use *,
var str = "52.) Ken".replace(/\d+\.\)\s*/g,"");
console.log(str); //Ken
Dismantling regex used,
/ states regex left border
\d d states normal character d, if we want to make it match
numbers then we have to escape it with \
+ It states that one or more number should be there.
\. Again . is a metacharacter to match any valid character, so
escape it.
\) Parenthesis is also a metacharacter to close a group, escape
it.
\s* 12.) can be followed by zero or more spaces.
/ states regex right boundary.
g global flag, which used to do a search recursively.

You can do it like this
var testURL = "182.) Hailey Quen";
var output = testURL.substring(testURL.lastIndexOf(")") + 1).trim();
console.log(output);
*trim function will help to remove extra space if any.Hope it will help

Javascript split function not correct worked with specific regex

I have a problem. I have a string - "\,str\,i,ing" and i need to split by comma before which not have slash. For my string - ["\,str\,i", "ing"]. I'm use next regex
myString.split("[^\],", 2)
but it's doesn't worked.

Well, this is ridiculous to avoid the lack of lookbehind but seems to get the correct result.
"\\,str\\,i,ing".split('').reverse().join('').split(/,(?=[^\\])/).map(function(a){
return a.split('').reverse().join('');
}).reverse();
//=> ["\,str\,i", "ing"]

Not sure about your expected output but you are specifying string not a regex, use:
var arr = "\,str\,i,ing".split(/[^\\],/, 2);
console.log(arr);
To split using regex, wrap your regex in /..../

This is not easily possible with js, because it does not support lookbehind. Even if you'd use a real regex, it would eat the last character:
> "xyz\\,xyz,xyz".split(/[^\\],/, 2)
["xyz\\,xy", "xyz"]
If you don't want the z to be eaten, I'd suggest:
var str = "....";
return str.split(",").reduce(function(res, part) {
var l = res.length;
if (l && res[l-1].substr(-1) == "\\" || l<2)
// ^ ^^ ^
// not the first was escaped limit
res[l-1] += ","+part;
else
res.push(part);
return;
}, []);

Reading between the lines, it looks like you want to split a string by , characters that are not preceded by \ characters.
It would be really great if JavaScript had a regular expression lookbehind (and negative lookbehind) pattern, but unfortunately it does not. What it does have is a lookahead ((?=) )and negative lookahead ((?!)) pattern. Make sure to review the documentation.
You can use these as a lookbehind if you reverse the string:
var str,
reverseStr,
arr,
reverseArr;
//don't forget to escape your backslashes
str = '\\,str\\,i,ing';
//reverse your string
reverseStr = str.split('').reverse().join('');
//split the array on `,`s that aren't followed by `\`
reverseArr = reverseStr.split(/,(?!\\)/);
//reverse the reversed array, and reverse each string in the array
arr = reverseArr.reverse().map(function (val) {
return val.split('').reverse().join('');
});

You picked a tough character to match- a forward slash preceding a comma is apt to disappear while you pass it around in a string, since '\,'==','...
var s= 'My dog, the one with two \\, blue \\,eyes, is asleep.';
var a= [], M, rx=/(\\?),/g;
while((M= rx.exec(s))!= null){
if(M[1]) continue;
a.push(s.substring(0, rx.lastIndex-1));
s= s.substring(rx.lastIndex);
rx.lastIndex= 0;
};
a.push(s);
/* returned value: (Array)
My dog
the one with two \, blue \,eyes
is asleep.
*/

Find something which will not be present in your original string, say "###". Replace "\\," with it. Split the resulting string by ",". Replace "###" back with "\\,".
Something like this:
<script type="text/javascript">
var s1 = "\\,str\\,i,ing";
var s2 = s1.replace(/\\,/g,"###");
console.log(s2);
var s3 = s2.split(",");
for (var i=0;i<s3.length;i++)
{
s3[i] = s3[i].replace(/###/g,"\\,");
}
console.log(s3);
</script>
See JSFiddle

Javascript regex - remove all special characters except semi colon

In javascript, how do I remove all special characters from the string except the semi-colon?
sample string: ABC/D A.b.c.;Qwerty
should return: ABCDAbc;Qwerty

You can use a regex that removes anything that isn't an alpha character or a semicolon like this /[^A-Za-z;]/g.
const str = "ABC/D A.b.c.;Qwerty";
const result = str.replace(/[^A-Za-z;]/g, "");
console.log(result);

var str = "ABC/D A.b.c.;Qwerty";
var result = str.replace(/[^A-Za-z;]/g, "");​​ // 21ABCDAbc;Qwerty
Live DEMO

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;
}

How do I match a number inside square brackets with regex

I wrote a regular expression which I expect should work but it doesn't.
var regex = new RegExp('(?<=\[)[0-9]+(?=\])')
JavaScript is giving me the error：
Invalid regular expression :(/(?<=[)[0-9]+(?=])/): Invalid group
Does JavaScript not support lookahead or lookbehind?

This should work:
var regex = /\[[0-9]+\]/;
edit: with a grouping operator to target just the number:
var regex = /\[([0-9]+)\]/;
With this expression, you could do something like this:
var matches = someStringVar.match(regex);
if (null != matches) {
var num = matches[1];
}

Lookahead is supported, but not lookbehind. You can get close, with a bit of trickery.

To increment multiple numbers in the form of lets say:
var str = '/a/b/[123]/c/[4567]/[2]/69';
Try:
str.replace(/\[(\d+)\]/g, function(m, p1){
return '['+(p1*1+1)+']' }
)
//Gives you => '/a/b/[124]/c/[4568]/[3]/69'

If you're quoting a RegExp, watch out for double escaping your backslashes.