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!
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Andrew
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2 Answers
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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.
David Tang
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That's what I was going for. Much nicer and simpler than mine. – Demian Brecht Mar 29 '11 at 22:21
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1Can be simplified (and sped up) even more: `str.match(/v=([^&]+))[1];` – ridgerunner Mar 30 '11 at 05:31
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To get the first capture group () from the URL that matches v=***********:
url.match(/v=(.{11})/)[1]
mVChr
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Fixed it to just get the next 11 characters, so you have to assume that v= will never be blank. – mVChr Mar 29 '11 at 22:12