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On Dec 5, 2008, at 10:34 AM, Dermot wrote:
I was looking to test title1 and return that if it was defined else,
test title2 and return that if defined, failing that, return oldtitle.
The ? : operator won't do that. It tests title1 and if true returns
title2. That would not be what I was after!
I think we ended up with the same solution. :-) here what I had.
my $title;
if (defined($image->title1)) {
$title = $image->title1;
}
elsif (defined($image->title2)) {
$title = $image->title2;
}
else {
$title = $image->oldtitle;
}
Wouldn't this work for what you are trying to do:
my $title = $image->title1 || $image->title2 || $image->oldtitle;
Sincerely,
James Moser
[EMAIL PROTECTED]
There are 10 kinds of people in the world.
Those who understand binary, and those who do not.
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