Remove specific multiple ending characters on a string

Question

I want to remove certain characters when they are at the end of a string. I have a working regex but the issue is the following:

My code:

var text = "some string";
text.replace(/\,$|:$|;$|-$|\($|–$|#$|\.$/, '');

So let's see the issue

If text = "that a nice (", then my code outputs "that a nice ", which is exactly what I want.

But if if text has multiple matches such as

text = "that a nice ( #", then the output of my current code is: "that a nice ( "... but I want to remove also the ( when it's at the end of the string as you can see on my regexp. I'd need to trim it so that the white space is removed but then how to remove the ( again...

The question is how to remove any unwanted characters and make sure that the new output also does not include these characters at their (new) end ? I could have 2, 3 or whatever number of unwanted characters at the end

Possible duplicate of stackoverflow.com/questions/1144783/… — Kévin Bibollet
– Kévin Bibollet, Commented Mar 11, 2019 at 9:40
@iArcadia no my question is not about removing ALL occurences of a substring but remove them when they're at the END of a string (or at the end just before a white space at the end) — Mathieu
– Mathieu, Commented Mar 11, 2019 at 9:42
@anubhava no it does not work. "that a nice ( #"".replace(/[-,:;(#.–]+$/, '') is outputting "that a nice ( ", which is alreayd what i get but i don't want any trailing ( — Mathieu
– Mathieu, Commented Mar 11, 2019 at 9:44

Michał Turczyn · Accepted Answer · 2019-03-11 09:43:13Z

1

You could use pattern: (\s*[#(])+$

\s*[#(] matches zero or more spaces, then one or more one of characters in square brackets, where you should list all unwanted characters. And then it matches this whole pattern on ro more times. $ matches end of string to assure that we are at the end of a string.

Demo

answered Mar 11, 2019 at 9:43

Michał Turczyn

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10 Comments

Michał Turczyn Over a year ago

@Mathieu Then you should accept answer (green check mark on the left) and optionally upvote.

Mathieu Over a year ago

sure i'll do but i tried and SO says i have to wait minimum 5 minutes:)

anubhava Over a year ago

It seems OP wants to match many more special characters as attempted regex was: text.replace(/\,$|:$|;$|-$|\($|–$|#$|\.$/, '');

Michał Turczyn Over a year ago

@anubhava Sure. That's why I pointed "where you should list all unwanted characters" :)

Mathieu Over a year ago

@MichałTurczyn i do get an error though "Invalid regular expression: /(\s*[,:;-(])+$/: Range out of order in character class at <anonymous>:1:6" when one of the unwanted character i put is (. for ex: text.replace(/(\s*[,:;-(])+$/, '')

|

RK_15 · Accepted Answer · 2019-03-11 09:44:05Z

1

use this regex and try:

/(\s*[-,:;\(#\.])*$/

answered Mar 11, 2019 at 9:44

RK_15

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anubhava · Accepted Answer · 2019-03-11 09:45:44Z

1

You may use a character class with + quantifier that should include \s (whitespace):

var text = "that a nice ( #";
text = text.replace(/[-,\s(){}:;(#.–]+$/, '');
console.log(text);

answered Mar 11, 2019 at 9:45

anubhava

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Toto · Accepted Answer · 2019-03-11 09:50:00Z

0

This will remove all unwanted character at the end and keep a space.

var test = [
    "that a nice (",
    "that a nice ( #",
];
console.log(test.map(function (a) {
  return a.replace(/[-,:;(#.–](?:\s*[-,:;(#.–])*$/, '');
}));

answered Mar 11, 2019 at 9:50

Toto

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jo_va · Accepted Answer · 2019-03-11 09:52:40Z

0

Use a character group and match it any times at the end /[(,:;\-–#\.\s]*$/:

const sanitize = text => text.replace(/[(,:;\-–#\.\s]*$/, '');

console.log(sanitize('this is a nice,:;-–#.'));
console.log(sanitize('this is a nice ( '));
console.log(sanitize('this is a nice ( #'));
console.log(sanitize('this is a nice (:- '));

answered Mar 11, 2019 at 9:52

jo_va

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