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# 1  
Old 12-07-2009
Problem with a regular expression
Hello!

I'm working with AWK, and i have this code:
Code:
/<LOOP_TIME>/,/<\/LOOP_TIME>/

I want that match every everything between <LOOP_TIME> and </LOOP_TIME>, but not if the line have a "#" before the tags.

Someone can help me?

Thanks!
# 2  
Old 12-07-2009
will this work for what you need?

Code:
/^<LOOP_TIME>/,/<\/LOOP_TIME>/

If not.... you can always use 2 filters. ie:

Code:
grep -v ^\# file | nawk '/<LOOP_TIME>/,/<\/LOOP_TIME>'

# 3  
Old 12-07-2009
I can't use this because i'm working in an awk script
# 4  
Old 12-07-2009
E.g.
Code:
'/^[^#]*<LOOP_TIME>/,/^[^#]*<\/LOOP_TIME>/'

This won't help you is the # character is within quotes, but it is a start.
Last edited by Scrutinizer; 12-07-2009 at 06:01 PM..
# 5  
Old 12-07-2009
if i have:
Code:
#<LOOP_TIME>
    something
</LOOP_TIME>

it works, don't match

But i have:
Code:
<LOOP_TIME>
    something
#</LOOP_TIME>

it doesn't works, match when don't must
# 6  
Old 12-07-2009
Quote:
Originally Posted by claw82
if i have:
Code:
#<LOOP_TIME>
    something
</LOOP_TIME>

it works, don't match

But i have:
Code:
<LOOP_TIME>
    something
#</LOOP_TIME>

it doesn't works, match when don't must
Well that is the nature of the beast. The /pattern/,/pattern2/ construct will start when it reaches pattern and list all lines until it reaches either pattern2 or EOF
You could use something like this:
Code:
awk '/^[^#]*<LOOP_TIME>/{s=$0;next} {s=s"\n"$0} /^[^#]*<\/LOOP_TIME>/{print s}'

Last edited by Scrutinizer; 12-07-2009 at 07:27 PM..
# 7  
Old 12-08-2009
Ok , then

I use you solution with some modifications in my code and it works.

Thanks!
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