Pick out part of a url with javascript regex?

Question

If I have the following url:

http://www.youtube.com/watch?v=ysIzPF3BfpQ&feature=rec-LGOUT-exp_stronger_r2-2r-3-HM
or
http://www.youtube.com/watch?v=ysIzPF3BfpQ

How can I pick out just the 11 character string, ysIzPF3BfpQ?

Thanks for the help!

David Tang · Accepted Answer · 2011-03-29 22:17:34Z

4

str.match(/v=(.*?)(&|$)/)[1];

It looks for a v=, then the shortest string of characters (.*?), followed by either a & or the end of the string. The [1] retrieves the first grouping, giving: ysIzPF3BfpQ.

edited Mar 29, 2011 at 22:17

answered Mar 29, 2011 at 22:12

David Tang

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2 Comments

Demian Brecht Over a year ago

That's what I was going for. Much nicer and simpler than mine.

ridgerunner Over a year ago

Can be simplified (and sped up) even more: str.match(/v=([^&]+))[1];

mVChr · Accepted Answer · 2011-03-29 22:11:08Z

1

To get the first capture group () from the URL that matches v=***********:

url.match(/v=(.{11})/)[1]

answered Mar 29, 2011 at 22:11

mVChr

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2 Comments

Andrew Over a year ago

What if the url is just: youtube.com/watch?v=ysIzPF3BfpQ ?

mVChr Over a year ago

Fixed it to just get the next 11 characters, so you have to assume that v= will never be blank.

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