Python Regular Expression Help

[debasishgang7]

Hi all,

I wanna extract a certain link from a web page using python regular expression.

The scenario is like this..

The code:

blah...
...
....
http://www.test.com/file.ext " style="top:0px;width:100%;"
....
blah
blah
blah

I wanna extract the url "http://www.test.com/file.ext" from the page using python regular expression.

Thanks in advance!

[snippsat]

Read this.
http://stackoverflow.com/questions/1732348/regex-match-open-tags-except-xhtml-self-contained-tags

So regex it not the right tool when it comes to html/xml.
There is a reason why parser excit,python has 2 very good lxml and BeautifulSoup.

from BeautifulSoup import BeautifulSoup

html = """\
<div class="test" src="http://www.test.com/file.ext" style="top:0px;width:100%;>"
"""

soup = BeautifulSoup(html)
tag = soup.find('div')
print tag['src']
#--> http://www.test.com/file.ext

So in a lager page your search would be more specific something like this.

from BeautifulSoup import BeautifulSoup

html = """\
<div class="test" src="http://www.test.com/file.ext" style="top:0px;width:100%;>"
"""

soup = BeautifulSoup(html)
tag = soup.findAll('div', {'class': 'test'})
print tag[0]['src']
#--> http://www.test.com/file.ext

[debasishgang7]

Well thanks for your suggestion,but in this case its not working.I am getting "IndexError: list index out of range" error.May be its because i am trying with huge page.And one more thing is the part of this html code is inactive means its between this tags.
I will be very thank full if you can solve this with a regular expression which will extract the url between

[pyTony]

What have you tried looks simple match betseen 'start and end tags'?

[Gribouillis]

Here is the solution.

[snippsat]

Well thanks for your suggestion,but in this case its not working.I am getting "IndexError: list index out of range" error.May be its because i am trying with huge page.And one more thing is the part of this html code is inactive means its between this tags.

That may be because you making and error,impossibile to say without seeing some code.
Regex ...no,but here something you can look at.

>>> import re
>>> re.findall(r'class="test" src="(.*?)"', html)
['http://www.test.com/file.ext']
>>> ''.join(re.findall(r'class="test" src="(.*?)"', html))
'http://www.test.com/file.ext'
>>>