On Jan 5, 2008 11:50 AM, Reese <[EMAIL PROTECTED]> wrote:
> Daniel Brown wrote:
>
> > Do you expect the value of $key in this condition to be a literal zero?
> > $twoyears = array('alphanumeric_code1', 'alphanumeric_code2',
> > 'alphanumeric_code3', 'alphanumeric_code4',
> > 'alphanumeric_code5',
> > 'alphanumeric_code6', 'alphanumeric_code7');
> > $key = in_array($sPromocode,$twoyears);
> > if($key=='0')
>
>
> I changed
>
> if($key=='0')
>
> to
>
> if(!isset($key=='1'))
>
> to see what effect that change might make, the server threw an error
> so I set it back to its original state:
>
> Parse error: parse error, unexpected T_IS_EQUAL, expecting ',' or ')' in
> /[PATH]/login.php on line 16
That's because isset() isn't able to eval() an expression. Remove
the !isset() part, or the =='1' part and that will remove the parse
error.
--
Daniel P. Brown
[Phone Numbers Go Here!]
[They're Hidden From View!]
If at first you don't succeed, stick to what you know best so that you
can make enough money to pay someone else to do it for you.
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