How to use Python regular expression to get Image src?

Question

How to use regular expression to get src of image from the following html string using Python

<td width="80" align="center" valign="top"><font style="font-size:85%;font-family:arial,sans-serif"><a href="http://news.google.com/news/url?sa=t&amp;fd=R&amp;usg=AFQjCNFqz8ZCIf6NjgPPiTd2LIrByKYLWA&amp;url=http://www.news.com.au/business/spain-victory-faces-market-test/story-fn7mjon9-1226390697278"><img src="//nt3.ggpht.com/news/tbn/380jt5xHH6l_FM/6.jpg" alt="" border="1" width="80" height="80" /><br /><font size="-2">NEWS.com.au</font></a></font></td>

I tried to use

matches = re.search('@src="([^"]+)"',text)
print(matches[0])

But got nothing

Answer 1

Instead of regex, you could consider using BeautifulSoup:

>>> from bs4 import BeautifulSoup
>>> soup = BeautifulSoup(junk)
>>> soup.findAll('img')
[<img src="//nt3.ggpht.com/news/tbn/380jt5xHH6l_FM/6.jpg" alt="" border="1" width="80" height="80" />]
>>> soup.findAll('img')[0]['src']
u'//nt3.ggpht.com/news/tbn/380jt5xHH6l_FM/6.jpg'