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StupidScript · 11-14-2006, 11:15 AM

You are using two kinds of checks in your control structure: isset and ==. I believe in this case the isset is what's causing your error whenever variable $i has not been initialized.

Code:

if($i && $i=="logout") {

would be the extended illustration of (perhaps) what you are trying to do. In practice,

Code:

if($i=="logout") {

would have the desired result. Not only must $i have been initialized, but it also must have the value "logout". If $i is neither, it won't match.

11-14-2006, 11:15 AM	#2 (permalink)
StupidScript Senior Member Join Date: 09-22-06 Location: Los Angeles Posts: 663 iTrader: 0 / 0% Latest Blog: None	You are using two kinds of checks in your control structure: isset and ==. I believe in this case the isset is what's causing your error whenever variable $i has not been initialized. Code: if($i && $i=="logout") { would be the extended illustration of (perhaps) what you are trying to do. In practice, Code: if($i=="logout") { would have the desired result. Not only must $i have been initialized, but it also must have the value "logout". If $i is neither, it won't match.