PHP Warning: Undefined variable:

[skd02]

How do you surpress the following PHP Warning:
Undefined variable: submitjoke in C:\Inetpub\wwwroot/php/deletejokes.php on line xx

Basically, there is an HTML form on the .php page but through the use of the PHP if: else: contol structure the PHP variable $submitjoke created by the NAME="submitjoke" form attribute is not processed at that time leading to the undefined variable.

I hope I was clear!

[Skunk]

It sounds like PHP is running in 'safe mode' of some sort - do you have control over the php.ini file? If so post it on here (or private message the file to me) and I'll point out what's wrong with it.

[freakysid]

Can you post the code? Is this from Kevin Yank's tutorial? I did Kev's tutorial earlier this year - and came accross some bugs that occured because of language changes between PHP3 (which he wrote the tutorial in) and PHP4.

For example:
if ("SUBMIT" == $submitjoke)
Causes an error in PHP4 if $submitjoke is not set

but:
if ($submitjoke == "SUBMIT")
Works regardless of whether $submitjoke existed prior to the conditioinal statement!