Re^2: trim leading & trailing whitespace

[thinker]

Hi bart

In the example I gave above,

 s/^\s*(.*?)\s*$/$1/gm

the combination of the \m modifier, and the line anchors ^ and $ will ensure the newlines are left alone

cheers

thinker

[bart]

$_ = "   no  \n   \n   \n    they    \n \n    won't     \n\n\n";
s/^\s*(.*?)\s*$/$1/gm;
print;

Result:

no
they
won't

All blank lines are gone.

[thinker]

Hi bart,

I see what you mean now. I thought you were suggesting it would remove the newline characters, not just the blank lines. Your way is more correct.

cheers

thinker

[tlm]

How about this:

$_ = "   yes  \n   \n   \n    they    \n \n    will     \n\n\n";
s/^\s*?(?:\b(.*?)\s*?)?(\n)?$/$1$2/gm;
print;
__END__
% perl 443391.pl
yes


they

will

[bart]

Danger, don't use \b if you want to detect the edges between spaces an nonspaces. It won't work for nonspace+nonword characters, like quotes and punctuation.

Perhaps replace the \b(.*?) with (\S.*?).

Still, you're not running this with warnings enabled, are you? Both $1 and $2 are optional, so both can be undef, but you use their value regardless. This runs without warnings:

s/^\s*?((?:\S.*?)?)\s*?(\n?)$/$1$2/gm;