String.replace regex is not replacing `*` in string

Question

I have a string that comes in and I am running a string.replace on it and it is not removing the * character from the string but it is removing the | character.

var s = "|*55.6";
s = s.replace(/[^0-9.]/, "");
console.log(s);

Possible duplicate of How to replace all occurrences of a string in JavaScript? — gyre
– gyre, Commented Mar 9, 2017 at 5:17

Community · Accepted Answer · 2017-05-23 11:46:46Z

4

Use a global regular expression (/.../g) if you want to replace more than a single match:

s = s.replace(/[^0-9.]+/g, "") //=> "55.6"

Edit: Following your pattern with + (which matches more than one of a pattern in succession) will produce fewer matches and thus make the replacement with "" run more efficiently than using the g flag alone. I would use them both together. Credit to Jai for using this technique in his answer before me.

Demo Snippet

var s = "|*55.6"
s = s.replace(/[^0-9.]+/g, "")
console.log(s) //=> "55.6"

edited May 23, 2017 at 11:46

CommunityBot

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answered Mar 9, 2017 at 5:13

gyre

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Comments

Sangbok Lee · Accepted Answer · 2017-03-09 05:14:02Z

1

It's because replace executed just once. Use g option to replace every match.

s = s.replace(/[^0-9.]/g, "");

answered Mar 9, 2017 at 5:14

Sangbok Lee

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Comments

bvrakvs · Accepted Answer · 2017-03-09 06:09:02Z

1

You haven't used a global flag for your regex, that is why the first occurrence was replaced, than the operation ended.

var s = "|*55.6";
s = s.replace(/[^\d.]/g, "");
console.log(s);

Using + quantifier alone wouldn't help if you have separate occurrences of [^\d.], though would be better to use together with g flag. For example;

var s = "|*55.6a"
s = s.replace(/[^0-9.]+/, "")
console.log(s) //=> "55.6a not OK

edited Mar 9, 2017 at 6:09

answered Mar 9, 2017 at 5:13

bvrakvs

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Comments

bvrakvs · Accepted Answer · 2018-11-07 10:17:22Z

1

You need to use a global (g) flag / modifier or a plus (+) quantifier when replacing the string;

var s = "|*55.6";
s = s.replace(/[^0-9.]/g, "");
console.log(s);

var s = "|*55.6";
s = s.replace(/[^0-9.]+/, "");
console.log(s);

edited Nov 7, 2018 at 10:17

bvrakvs

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answered Mar 9, 2017 at 5:17

m87

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Comments

bvrakvs · Accepted Answer · 2018-11-07 10:17:31Z

1

You are missing + to look for one or more occurrences:

var s = "|*55.6";
s = s.replace(/[^0-9.]+/, "");
console.log(s);

Also if you are interested to takeout the numbers, then you can use .match() method:

var s = "|*55.6";
s = s.match(/[(0-9.)]+/g)[0];
console.log(s);

edited Nov 7, 2018 at 10:17

bvrakvs

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answered Mar 9, 2017 at 5:14

Jai

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Comments

shizhz · Accepted Answer · 2017-03-09 05:44:31Z

0

Some answer mentioned the Quantifiers +, but it will still only replace the first match:

console.log("|*45.6abc".replace(/[^0-9.]+/, ""))

If you need to remove all non-digital chars(except .), the modifier/flag g is still required

console.log("|*45.6,34d".replace(/[^0-9.]+/g, ""));

answered Mar 9, 2017 at 5:44

shizhz

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