Selecting numbers only in a JavaScript string

Question

I want to select all the digits from a given string. I tried with the code below, but it does not return all the numbers in the string:

var match = /\d+/.exec("+12 (345)-678.90[]");
console.log(match.toString());

It only returns 12, while I expect it to return 1234567890.

You're only selecting one match. There are four matches in there. "+12 (345)-678.90[]".replace(/\D/g, ""); should do it — CodingIntrigue
– CodingIntrigue, Commented May 27, 2015 at 11:54

divakar · Accepted Answer · 2015-05-27 11:59:44Z

5

simple implementation will be

var value='+12 (345)-678.90[]'.replace(/\D+/g, '');
console.log(value);

answered May 27, 2015 at 11:59

divakar

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Comments

Satpal · Accepted Answer · 2015-05-27 11:58:25Z

2

You need to use global flag, it will return you an array of matched data the you can use join() it.

"+12 (345)-678.90[]".match(/\d+/g).join('');

alert("+12 (345)-678.90[]".match(/\d+/g).join(''))

answered May 27, 2015 at 11:58

Satpal

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Comments

moonwave99 · Accepted Answer · 2015-05-27 11:58:00Z

1

Use the global flag:

"+12 (345)-678.90[]".match(/\d+/g)

answered May 27, 2015 at 11:58

moonwave99

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1 Comment

Khadim Ali Over a year ago

This was returning the matches with , .. like 12,345,678,90. Adding .join('') returned perfectly was required. :)

Wiktor Stribiżew · Accepted Answer · 2015-05-27 12:10:13Z

1

The \d+ pattern will return consecutive digits only, and since you running exec once without g option, it will only give you the first occurrence of consecutive digits.

Use this:

var re = /\d+/g; 
var str = '+12 (345)-678.90[]';
var res = "";
while ((m = re.exec(str)) !== null) {
    res += m[0];
}
alert(res);

Output is 1234567890, as we append found digit sequences to the res variable.

edited May 27, 2015 at 12:10

answered May 27, 2015 at 11:58

Wiktor Stribiżew

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1 Comment

Wiktor Stribiżew Over a year ago

@Juhana: Good catch, looks like it is redundant here. There are cases when we need to count matches, and then it is useful.

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