Regex replace with number after capture group?

Question

I have a regex pattern like this:

([0-9]*)xyz

I wish to do substitution like this:

$10xyz

The problem is that the $1 is a capture group and the 0 is just a number I want to put into the substitution. But regex thinks I'm asking for capture group $10 instead of $1 and then a zero after it.

How do I reference a capture group and immediately follow it with a number?

Using JavaScript in this case.

UPDATE As pointed out below, my code did work fine. The regex tester I was using was accidentally set to PCRE instead of JavaScript.

This is addressed by the following entry in the [Stack Overflow Regular Expressions FAQ](http://stackoverflow.com/a/22944075/2736496), under "Groups": *[How to follow a numbered capture group, such as \1, with a number?](http://stackoverflow.com/q/5984633/3622940)* — aliteralmind, Jan 02 '15 at 17:07

hwnd · Accepted Answer · 2015-01-02T18:27:33.720

5

Your code indeed works just fine. In JavaScript regular expression replacement syntax $10 references capturing group 10. However, if group 10 has not been set, group 1 gets inserted then the literal 0 afterwards.

var r = '123xyz'.replace(/([0-9]*)xyz/, '$10xyz');
console.log(r); //=> "1230xyz"

edited Jan 02 '15 at 18:27

answered Jan 02 '15 at 17:34

hwnd

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You are correct. Sorry, the tester I was using was set to use PCRE instead of JS. – Jake Wilson Jan 02 '15 at 19:06

score 1 · Answer 2 · answered Jan 02 '15 at 17:17

1

Your code does work unless I'm missing something:

var str = "3xyz";
var str1 = str.replace(/([0-9]*)xyz/, "$10xyz");
alert(str1); // alerts 30xyz

answered Jan 02 '15 at 17:17

eatonphil

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You are correct. Sorry, the tester I was using was set to use PCRE instead of JS. – Jake Wilson Jan 02 '15 at 17:48

Regex replace with number after capture group?

2 Answers2