in reply to scalar out of a loop

Maybe you don't want to export a variable out of the loop, but set an existing variable to a different value?

my $comment= '- no comment found -';
                       foreach my $hash_ref2 (@result2) {
                               $comment = $hash_ref2->{'comment'};
                               if ($comment =~ /^this is an automatic 
+comment/) {
                                       last;
                               }
                       }
print $comment;

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Re^2: scalar out of a loop
by mitchreward (Acolyte) on Jul 21, 2014 at 14:24 UTC
    No, but you're right that was not complete. I've updated my post.
[reply]

      My answer is the same, except that you'd replace my $comment by my $done.

      Or maybe I misunderstood your question still.

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