Python regex look-behind requires fixed-width pattern

Question

Hi

When trying to extract the title of a html-page I have always used the following regex:

(?<=<title.*>)([\s\S]*)(?=</title>)

Which will extract everything between the tags in a document and ignore the tags themselves. However, when trying to use this regex in Python it raises the following Exception:

Traceback (most recent call last):  
File "test.py", line 21, in <module>
    pattern = re.compile('(?<=<title.*>)([\s\S]*)(?=</title>)')
File "C:\Python31\lib\re.py", line 205, in compile
    return _compile(pattern, flags)   
File "C:\Python31\lib\re.py", line 273, in _compile
    p = sre_compile.compile(pattern, flags)   File
"C:\Python31\lib\sre_compile.py", line 495, in compile
    code = _code(p, flags)   File "C:\Python31\lib\sre_compile.py", line 480, in _code
_compile(code, p.data, flags)   File "C:\Python31\lib\sre_compile.py", line 115, in _compile
    raise error("look-behind requires fixed-width pattern")
sre_constants.error: look-behind requires fixed-width pattern

The code I am using is:

pattern = re.compile('(?<=<title.*>)([\s\S]*)(?=</title>)')
m = pattern.search(f)

if I do some minimal adjustments it works:

pattern = re.compile('(?<=<title>)([\s\S]*)(?=</title>)')
m = pattern.search(f)

This will, however, not take into account potential html titles that for some reason have attributes or similar.

Anyone know a good workaround for this issue? Any tips are appreciated.

Answer 1

Toss out the idea of parsing HTML with regular expressions and use an actual HTML parsing library instead. After a quick search I found this one. It's a much safer way to extract information from an HTML file.

Remember, HTML is not a regular language so regular expressions are fundamentally the wrong tool for extracting information from it.

Answer 2

Here's a famous answer on parsing html with regular expressions that does a great job of saying, "don't use regex to parse html."

Answer 3

If you just want to get the title tag,

html=urllib2.urlopen("http://somewhere").read()
for item in html.split("</title>"):
    if "<title>" in item:
        print item[ item.find("<title>")+7: ]

Answer 4

What about something like:

 r = re.compile("(<title.*>)([\s\S]*)(</title>)")
 title = r.search(page).group(2)