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perl: conditionaly remove space

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# 1  
Old 10-14-2009
perl: conditionaly remove space

I want to remove all space NOT starting with http, the following works, but not elegant, is there better way?

$echo " a http  b" | perl -lne 's/\shttp/#http/g; s/\s+//g;  s/#http/ http/g; print $_;'
a httpb

# 2  
Old 10-15-2009
Not able to understand exactly, but this may be the thing you want..

echo "a http b" | perl -lne 's/(http)\s+/$1/g; print'

# 3  
Old 10-15-2009
The shortest I could get.
$ echo " a http  b" | perl -ple 's/\s(?!http)//g'
a httpb

# 4  
Old 10-15-2009
Originally Posted by pludi
The shortest I could get.
$ echo " a http  b" | perl -ple 's/\s(?!http)//g'
a httpb

great that is what I need delete all space except those before http.

Can you explain the usage of "?!", I googled but still couldn't understand it.

Got it:

A zero-width negative lookahead assertion. For example /foo(?!bar)/ matches any occurrence of ``foo'' that isn't followed by ``bar''. Note however that lookahead and lookbehind are NOT the same thing. You cannot use this for lookbehind.
If you are looking for a ``bar'' that isn't preceded by a ``foo'', /(?!foo)bar/ will not do what you want. That's because the (?!foo) is just saying that the next thing cannot be ``foo''--and it's not, it's a ``bar'', so ``foobar'' will match. You would have to do something like /(?!foo) for that. We say ``like'' because there's the case of your ``bar'' not having three characters before it. You could cover that this way: /(?:(?!foo)...|^.{0,2})bar/. Sometimes it's still easier just to say:

if (/bar/ && $` !~ /foo$/)

For lookbehind see below.

Last edited by pludi; 10-16-2009 at 02:26 AM..
# 5  
Old 10-16-2009
Congratulations on finding the answer. Zero-width assertions are a very valuable tool sometimes. Sadly, they're only available in Perl (except if someone can correct me on that).

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