regexp to replace string after third '-' in javascript

Question

I need to replace a string only after 3 occurences of '-'. Can someone give me a pattern for it?

ex: xxyy-xxyy-xxyy-xxyy where xx can be more than one character or number and I need to replace the last set of characters.

finalString = String.replace(/xx$/m, "z")

and this doesn't work because I don't know the last set of characters.

Thanks in advance.

@PraveenJeganathan: That is for vim, will the regex in that answer work for Javascript as well? — Bill Lynch
– Bill Lynch, Commented Jan 29, 2014 at 15:04
@anubhava Ohh I missed it. I have retracted my close votes. Thanks. — Praveen
– Praveen, Commented Jan 30, 2014 at 3:37

anubhava · Accepted Answer · 2014-01-29 15:03:52Z

1

You can do:

var s = 'xxyy-xxyy-xxyy-xxyy';
var r = s.replace(/^((?:[^-]*-){3}).*/, '$1foo');
//=> xxyy-xxyy-xxyy-foo

answered Jan 29, 2014 at 15:03

anubhava

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2 Comments

JayJay Over a year ago

First of all, thanks. This is almost perfect but I don't want to replace the whole set of characters just the XX part. //=> xxyy-xxyy-xxyy-fooyy

JayJay Over a year ago

Another tip: this should search for the 'xx' after the last '-' (last '-' or third '-' because I that the string will only have 3 '-').

Toto · Accepted Answer · 2014-01-29 15:29:10Z

0

How about:

finalString = String.replace(/xx([^-]+)$/m, "z$1")

edited Jan 29, 2014 at 15:29

answered Jan 29, 2014 at 15:06

Toto

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3 Comments

JayJay Over a year ago

Thanks and again, it's almost there. it replaces everything after the last '-' but I need //=> xxyy-xxyy-xxyy-fooyy

JayJay Over a year ago

No, I will try to explain me better. After the last '-' there will be a set of characters. I know that will start by 'xx' (this is what I want to replace) but after there will be more characters that I want to remain there and I don't know which characters are. Is it clear?

Toto Over a year ago

@JayJay: Ok, I see. Have a try with my modified answer, it'll replace last xx by z, keeping the rest unchanged.

Suganthan Madhavan Pillai · Accepted Answer · 2014-01-29 15:35:26Z

0

This is something i took from anubhava's answer, hope this is what you required

var s = 'xxyy-xxyy-xxyy-xxyy';
var r = s.replace(/^((?:[^-]*-){3})xx*/, '$1foo');

answered Jan 29, 2014 at 15:35

Suganthan Madhavan Pillai

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1 Comment

JayJay Over a year ago

Athough this question is already answered I confirm that this solution works as well. Thanks.

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