How to find everything which goes after symols #TR= and it is inside [ ] using re module. For example #TR=[ dfgg dfgddfg dgfgf dgdgdg dfgfg ]
3 Answers
import re txt = '#TR=[ dfgg ] a kuku #TR=[ala ma kota]' If you want to search for just the first occurrence of this pattern, use:
matches = re.search('#TR=\[([^\]]*)\]', txt) if matches: print(repr(matches.group(1))) ' dfgg dfg ' If you want to find all occurrences in the text, use:
matches = re.findall('#TR=\[([^\]]*)\]', txt) if matches: print(matches) [' dfgg ', 'ala ma kota'] Remember to check whether the characters you are searching for have special meaning in regular expressions (like [ or ]). If they are special, escape them with the backslash: \[.
Also remember, that by default, regular expressions are "greedy" which means they try to get as much text to match the pattern as possible; so if you use .* (which means "match any character except newline"; details) instead of [^\]]* (which means "match until the ] is found, and stop before it"), too much text could be matched:
matches = re.findall('#TR=\[(.*)\]', txt) if matches: print(matches) [' dfgg ] a kuku #TR=[ala ma kota'] You can also use the "non-greedy" modifier ? in your pattern, after the qualifier (*, +) which enables the "the-less-characters-the-better" matching (use *?, +?). The result could be more readable:
'#TR=\[(.*?)\]' instead of:
'#TR=\[([^\]]*)\]' There's a great online tool to test your patterns as-you-type: RegExr by Grant Skinner.
Comments
import re # compile the regex exp = re.compile('.*\[(.*)\].*') txt = r"#TR=[ dfgg dfgddfg dgfgf dgdgdg dfgfg ]" match = exp.match(txt) # grab the text between the square brackets result = match.group(1) 
1 Comment
Sasha Chedygov
Sorry, I edited your answer by mistake, meant to edit my own. Reverted my change.
(?<=#TR=[)[^]]*(?=])
[ ]inside your top level of[ ]?