----Original Message-----
From: Ho, Tony [mailto:[EMAIL PROTECTED]]
Subject: RE: "Use of uninitialized value" error message
Hi Nikola/Jason
Thanks for the help.
The variable $result_value1 is declared within the subroutine and is never
used outside the subroutine.
I originally had the following piece of code (which I forgot to show you
guys in the previous email):
$result_value1 = $database1{$input_key1}
But I changed the above code to the following and it worked ("Use of
uninitialized value" error message" did not appear):
$result_value1 = $database1{$input_key1} || " ";
------------------------------------
ahh... that would do it, though I don't think that's the preferred way to
solve your problem.
You are getting the "Use of uninitialized value" warning message because
$database1{$input_key1} is undef. You are simply bypassing that by making
$result_value1 a " " if it would otherwise return undef. By using your
original code ( $result_value1 = $database1{$input_key1}; ) with "if
($result_value1);" you do what you want and avoid the warning because you
are now checking to see if $result_value1 is defined. I know that Nikola
says you will still get the warning, but it always works for me... :)
Hope this helps...
Jason
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