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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);
gyre
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jshill103
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    Possible duplicate of [How to replace all occurrences of a string in JavaScript?](http://stackoverflow.com/questions/1144783/how-to-replace-all-occurrences-of-a-string-in-javascript) – gyre Mar 09 '17 at 05:17

6 Answers6

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"
Community
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gyre
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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
buræquete
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1

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

s = s.replace(/[^0-9.]/g, "");
Sangbok Lee
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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);
buræquete
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Jai
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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);
buræquete
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m87
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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, ""));
shizhz
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