How do I ensure that re.findall() stops at the right place?

问题

Here is the code I have:

a='<title>aaa</title><title>aaa2</title><title>aaa3</title>'
import re
re.findall(r'<(title)>(.*)<(/title)>', a)

The result is:

[('title', 'aaa</title><title>aaa2</title><title>aaa3', '/title')]

If I ever designed a crawler to get me titles of web sites, I might end up with something like this rather than a title for the web site.

My question is, how do I limit findall to a single <title></title>?

回答1:

Use re.search instead of re.findall if you only want one match:

>>> s = '<title>aaa</title><title>aaa2</title><title>aaa3</title>'
>>> import re
>>> re.search('<title>(.*?)</title>', s).group(1)
'aaa'

If you wanted all tags, then you should consider changing it to be non-greedy (ie - .*?):

print re.findall(r'<title>(.*?)</title>', s)
# ['aaa', 'aaa2', 'aaa3']     

But really consider using BeautifulSoup or lxml or similar to parse HTML.

回答2:

Use a non-greedy search instead:

r'<(title)>(.*?)<(/title)>'

The question-mark says to match as few characters as possible. Now your findall() will return each of the results you want.

http://docs.python.org/2/howto/regex.html#greedy-versus-non-greedy

回答3:

re.findall(r'<(title)>(.*?)<(/title)>', a)

Add a ? after the *, so it will be non-greedy.

回答4:

It will be much easier using BeautifulSoup module.

https://pypi.python.org/pypi/beautifulsoup4

来源：https://stackoverflow.com/questions/17765805/how-do-i-ensure-that-re-findall-stops-at-the-right-place