I started playing around with Python (2.6.6) trying to parse a file. No matter what I tried so far I am not able to print only the lines/group/backreference that have been affected by the substitution. Instead I get also the other lines too
File to parse:
Code:
[serverClass:One] some more text
config = alpha.cfg
[serverClass:Two] some more text
config = bravo.cfg
[serverClass:Three] some more text
config = charly.cfg
Current code:
Code:
import re
f = open('infile', 'r')
lines = f.readlines()
for line in lines:
x = re.sub(r"^\s*\[serverClass:([^]]*)\] .*$", r"\1", line.rstrip())
print(x)
I assume there is something needed like the behavior of sed's -n and /p to basically print nothing but the lines that have been matched.
Output:
Code:
$ python ./mach.py
One
config = alpha.cfg
Two
config = bravo.cfg
Three
config = charly.cfg
Expected output:
Code:
One
Two
Three
I saw that there are more functions/methods in the module re like re.group etc. but I would like to know why it doesn't work with my example.
Thanks in forward for a hint with this rather basic problem.
import re
f = open('infile', 'r')
lines = f.readlines()
for line in lines:
if re.match('^.*:(.*)].*$',line.rstrip()):
x = re.sub(r'^.*:(.*)].*$', r'\1', line.rstrip())
print x
Thank you, itkamaraj. I had it tried so far with match() and search() and it didn't work.
Now I just noticed why - I used:
Code:
re.match('<partial-pattern>', ...
...where I assumed I can use just a part of a pattern like when you use Linux/Unix grep; this seems not working so I have to use at least something like:
Code:
re.match('.*<partial-pattern>.*', ...
cheers
zaxxon
Last edited by zaxxon; 09-01-2016 at 09:24 AM..
Reason: clarification, typos
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09-01-2016
