Extract only the link from string using regex

Question

I want to extract link from this below mentioned string.

 str = /url?q=http://www.example.com/services/blog/first-article&sa=U&ei...

I used the following regular expression to get that link.But it fetches the full url after "http" because I mentioned the pattern to be.What I want is to get only URL before the pattern "&sa" (ie) "http://www.example.com/services/blog/first-article"

 links = re.findall(r'/url\?q=(http://.*)', str)
 print links  # http:example.com/services/blog/first-article&sa=U&ei...

Why not r'/url\?q=(http://.*?)&sa=.*'?

RedX
– RedX

2014-02-25 09:37:55 +00:00
Commented Feb 25, 2014 at 9:37 — RedX
– RedX, Commented Feb 25, 2014 at 9:37

isedev · Accepted Answer · 2014-02-25 09:38:42Z

2

This is the regular expression you need:

links = re.findall(r'/url\?q=(http://[^&]*)', str)

In words: get everything after /url?q=, starting with http:// and which doesn't contain a & character.

answered Feb 25, 2014 at 9:38

isedev

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