2008/12/5 Raymond Wan <[EMAIL PROTECTED]>:
>
> Hi Dermot,
>
>
> Dermot wrote:
>> Sorry. Someone on the list mentioned to me before, a reference is
>> always true, so the test won't work.
>>
>
> Maybe I have missed something, but it sounds like you want something like
> [NB: This is not Perl code, but sort of pseudocode]:
>
> if (defined (title1)) {
> $title = title1;
> }
> elsif (defined (title2)) {
> $title = title2;
> }
> else {
> $title = oldtitle;
> }
>
> Is this correct? If so, the ? operator is not what you want. What the
> operator does is this. For "a ? b : c" it does the test on "a" but not on
> "b". All it does is say, "Is 'a' true? If so, return 'b'. Otherwise,
> return 'c'." You aren't doing any test on b, but it sounds like you want
> to? (Sorry, you showed your table, but you didn't quite explain what
> title1, title2, and oldtitle means -- so, I am guessing here that you also
> want to test title2? I am also assuming that you do not want either title1
> or title2, but would prefer title1 -- hence the order of the pseudocode
> above.)
Your right.
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;
}
Thanx Raymond,
Dp.
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