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?
$twoyears = array('alphanumeric_code1', 'alphanumeric_code2',
'alphanumeric_code3', 'alphanumeric_code4',
Daniel Brown wrote:
if(!isset($key=='1')) //caused parse error
That's because isset() isn't able to eval() an expression.
Got it, I see the mistake now.
Remove the !isset() part, or the =='1' part and that will remove
the parse error.
I changed it to if(!isset($key)) and
Greetings,
I've been lurking for several weeks, I thought I'd post to describe
a problem I've been having in the hope that a solution can be found.
And my thanks to Casey, for his offlist assistance with another,
unrelated issue earlier this week. :-)
I apologize up front, for what is probably
On Jan 4, 2008 9:54 AM, Reese [EMAIL PROTECTED] wrote:
Greetings,
I've been lurking for several weeks, I thought I'd post to describe
a problem I've been having in the hope that a solution can be found.
And my thanks to Casey, for his offlist assistance with another,
unrelated issue earlier
Web Design Company wrote:
Someone?
Me31!1!1ONE
Please, if you do not need amplifying information or if you do
not intend to pose a suggestion, it is better to remain silent.
I wasn't helped by your Someone? post, no one else was either.
Reese
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Daniel Brown wrote:
[snip=all]
Reese,
While I noticed several areas for improvement in the code (such as
being sure to exit; after calling header(Location: ); ), two
things primarily come to mind:
Do you expect the value of $key in this condition to be a literal zero?
On Jan 4, 2008 11:55 AM, Reese [EMAIL PROTECTED] wrote:
Web Design Company wrote:
Someone?
Me31!1!1ONE
Please, if you do not need amplifying information or if you do
not intend to pose a suggestion, it is better to remain silent.
I wasn't helped by your Someone? post, no one else was
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