Posted by Jerry Stuckle on 08/18/06 01:28

mens libertina wrote:
> Jerry Stuckle wrote:
>
>>It wouldn't. $info is local to this function. You need to have a
>>variable with global scope - or at least one which is available to the
>>main code, not just a function.
>>
>
>
> After I read your post, I declared $info as "$info = array();" at the
> top of the program with the same result. Did I misunderstand what you
> meant by global scope?
>
> Thanks.
>

But now your problem is $info in the function is not the same as $info
in the main program.

When declared as a parameter like this, $info is only local to the
function. Any other $info is completely separate from the one in the
function.


--
==================
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Jerry Stuckle
JDS Computer Training Corp.
jstucklex@attglobal.net
==================

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