CODEWITHSUNDEEP
pythonregex
joshft91
joshft91

Reputation: 1885

Match string between special characters

I've messed around with regex a little bit but am pretty unfamiliar with it for the most part. The string will in the format:

\n\n*text here, can be any spaces, etc. etc.*

The string that I will get will have two line breaks, followed by an asterisk, followed by text, and then ending with another asterisk.

I want to exclude the beginning \n\n from the returned text. This is the pattern that I've come up with so far and it seems to work:

pattern = "(?<=\\n\\n)\*(.*)(\*)"

match = re.search(pattern, string)
if match:
    text = match.group()
    print (text)
else:
    print ("Nothing")

I'm wondering if there is a better way to go about matching this pattern or if the way I'm handling it is okay.

Thanks.

Upvotes: 0

Views: 205

Answers (2)

Casimir et Hippolyte
Casimir et Hippolyte

Reputation: 89547

You can avoid capturing groups and have the whole match as result using:

pattern = r'(?<=\n\n\*)[^*]*(?=\*)'

Example:

import re
print re.findall(r'(?<=\n\n\*)[^*]*(?=\*)','\n\n*text here, can be any spaces, etc. etc.*')

If you want to include the asterisk in the result you can use instead:

pattern = r'(?<=\n\n)\*[^*]*\*'

Upvotes: 2

bgporter
bgporter

Reputation: 36474

Regular expressions are overkill in a case like this -- if the delimiters are always static and at the head/tail of the string:

>>> s = "\n\n*text here, can be any spaces, etc. etc.*"
>>> def CheckString(s):
...     if s.startswith("\n\n*") and s.endswith("*"):
...         return s[3:-1]
...     else:
...         return "(nothing)"
>>> CheckString(s)
'text here, can be any spaces, etc. etc.'
>>> CheckString("no delimiters")
'(nothing)'

(adjusting the slice indexes as needed -- it wasn't clear to me if you want to keep the leading/trailing '*' characters. If you want to keep them, change the slice to 

return s[2:]

Upvotes: 1

Related Questions