Question

Sample text: String -> content within the rev tag (via lxml).

I m trying to remove the {{BLOCKS}} within the text.

I ve used the following regex to remove simple, one line blocks:

p = re.compile( {{*.*}} )
nonBracketedString = p.sub(  , bracketedString)

However this does not remove the first multi line bracketed section at the beginning of the content. How can one remove the multi-line, curly bracketed blocks?

EDIT:

Solution from answer:

p = re.compile( {{*?.*?}} , re.DOTALL)
nonBracketedString = p.sub(  , bracketedString)

Answer 1

Set the dotall flag.

p = re.compile( {{*.*?}} , re.DOTALL)
nonBracketedString = p.sub(  , bracketedString)

In the default mode, . matches any character except a newline. If the DOTALL flag has been specified, this matches any character including a newline.

http://docs.python.org/library/re.html

Also, you ll need a non-greedy match between the brackets: .*?

Answer 2

>>> import urllib2
>>> import re
>>> s = "".join(urllib2.urlopen( http://en.wikipedia.org/w/api.php?action=query&prop=revisions&titles=Italian%20War%20of%201542-1546&redirects&rvprop=content&format=xml ).readlines())
>>> p = re.compile( {{.*?}} , re.DOTALL)
>>> re.sub(p,   , s)
 <?xml version="1.0"?><api><query><redirects><r from="Italian War of 1542-1546" to="Italian War of 1542xe2x80x931546" /></redirects><pages><page pageid="3719774" ns="0" title="Italian War of 1542xe2x80x931546"><revisions><rev xml:space="preserve">



The    Italian War of 1542xe2x80x9346    was a conflict late in the [[Italian Wars]], ...

I ve truncated the output here, but there s enough to see that it s working.

Answer 3

Set the dotall flag-- this allows . to match newlines.

p = re.compile( {{*.*}} , re.DOTALL)
nonBracketedString = p.sub(  , bracketedString)

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