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
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'
pludi
October 15, 2009, 1:26am
3
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:http://www.perl.com/doc/manual/html/pod/perlre.html
\(?!pattern\)
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 /\(?!foo\)...bar/ 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.
pludi
October 16, 2009, 1:28am
5
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).