Python Regular Expression Matching: ## ##

Question

I\'m searching a file line by line for the occurrence of ##random_string##. It works except for the case of multiple #...

pattern=\'##(.*?)##\'
prog=re.compile(p         


        

Answer 1

Your problem is with your inner match. You use ., which matches any character that isn't a line end, and that means it matches # as well. So when it gets ###hey##, it matches (.*?) to #hey.

The easy solution is to exclude the # character from the matchable set:

prog = re.compile(r'##([^#]*)##')

Protip: Use raw strings (e.g. r'') for regular expressions so you don't have to go crazy with backslash escapes.

Trying to allow # inside the hashes will make things much more complicated.

EDIT: If you do not want to allow blank inner text (i.e. "####" shouldn't match with an inner text of ""), then change it to:

prog = re.compile(r'##([^#]+)##')

+ means "one or more."