> From: Brian Dessent
> Sent: Monday, October 20, 2003 10:43 PM
> Hannu E K Nevalainen wrote:
>
> > $ echo ' a b / c d e ' | \
sed -e 's/ *\(.*\) *\/ *\(.*\) */.\1.\2./'
> >
> > .a b .c d e .
> >
> > I want the output to be '.a b.c d e.' - that is; strip out the
> > trailing spaces.
> >
> > HOW do I achieve that? ( \s = any ws, \S = any non ws )
> >
> > Obviously \(.*\) grabs/includes the last space. My brain has stoppped
> > working, so right now I can't work around that :-I
>
> Personally, I can't stand the "basic" style regexps that sed uses, and
> I prefer perl-compatible. You can use the non-greedy modifier, for
> example:
>
> command | perl -pe 's!^ +(.*) +/ +(.*?) +$!.\1.\2.!'
Thanks a lot! Now all I have to do is understand where the difference is
;-7
$ echo ' a b / c d e ' | \
sed -re 's!^ +(.*) +/ +(.*?) +$!.\1.\2.!'
.a b.c d e.
It seems as the -r flag to sed made _this_ expression work right OOTB.
Hmm... lets see;
$ echo ' a b / c d e ' |
sed -e 's/^ \+\(.*\) \+\/ \+\(.*\) \+$/.\1.\2./'
.a b.c d e.
So THERE was the problem, some escaping needed.
As it seems my query wasn't that well formed... i.e. remove any leading
and/or trailing spaces on the parts. Parts separated by the slash.
This seems to do exactly what I'm after;
$ echo 'a b/c d e ' | \
sed -re 's- *(.*[^ ]) */ *(.*[^ ]) *$-.\1.\2.-'
Thanks for the input, Brian and Igor.
/Hannu E K Nevalainen, B.Sc. EE - 59?16.37'N, 17?12.60'E
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