Re: Less greedy pattern match

On 4/18/2008 2:50 AM, pk wrote:
> On Friday 18 April 2008 06:36, Ed Morton wrote:
> 
> 
>>Hang on, I spotted a couple more characters we could get rid of:
>>
>>   gawk 'NR%2' RS='</?t1>' ORS=
>>
>>So, it's now briefer than the perl solution:
>>
>>   perl -p0e 's%<t1>.+?</t1>%%gs'
>>
>>What does it do again ;-)?
> 
> 
> Ok ok, you won :-)
> Now let's just hope some perl diehard does not come up with a shorter
perl
> solution...
> 

I'm actually pretty impressed with the elegance of that solution (thanks
Rajan!)
and it's easily generalised to:

gawk -v RS='(begin|end)' -v ORS= 'NR%2'

where "begin" is the pattern that starts the segment you want to remove
(<t1> in
this case) and "end" is the pattern that terminates it (</t1>).

The only thing gawk-specific (or non-POSIX at least) about it now is the
use of
REs in the RS, but I'd like to see if there's a POSIX version anyway. This
is close:

awk -F'</?t1>' -v RS= '{for (i=1;i<=NF;i+=2) printf "%s",$i; print ""}'

but would add a trailing blank line at the end of the file, and won't
properly
handle blank lines between the "<tl>" and "</tl>". This would solve those
problems:

awk -F'</?t1>' 'BEGIN{RS=SUBSEP}{for (i=1;i<=NF;i+=2) printf "%s",$i;
print ""}'

but means the whole file has to be read in as one record which may not be
very
efficient for large files. Any ideas?

	Ed.