PHP error page problem - php

Question

I can't figure out why this script isn't working.
<?php
if (($_GET['p'] != 'index') &&
($_GET['p'] != 'reg') &&
($_GET['p'] != 'login') &&
($_GET['p'] != 'ad') &&
(!isset($_GET['p']))):
?>
<?php endif; ?>
I want to not display the error page if the $_GET is not set, which in my experience (!isset($_GET['p'])) should do.

Answer 1

If $_GET['p'] is not set, you can't check $_GET['p'] != 'index' and all the others. You'll have to check if it's set first:
<?php if(
! isset( $_GET['p'] ) ||
($_GET['p'] != 'index' &&
$_GET['p'] != 'reg' &&
$_GET['p'] != 'login' &&
$_GET['p'] != 'ad')
): ?>
A better solution would be to put all those values in an array, and check if $_GET['p'] is in the array:
<?php if(
! isset( $_GET['p'] ) ||
! in_array(
$_GET['p'],
array('index', 'reg', 'login', 'ad')
)
): ?>
EDIT:
Now that you provided some more info, here's what you should do:
if ( ! isset($_GET['p']) )
{
// We're at the index page, so don't display anything
}
else
{
if ( in_array( $_GET['p'], array('index', 'reg', 'login', 'ad') ) )
{
// Display your content window
}
else
{
// $_GET['p'] is not a valid value, display error
}
}

Answer 2

Your condition makes no sense. You're checking for 3 possible values of $_GET['p'] and then checking if $_GET['p'] is even set. Reverse your logic:
<?php
if(isset($_GET['p']))
{
// display error page
}
else
{
// do something else
}

Answer 3

You can check if $_GET['p'] is set by
if(isset($_GET['p']) {...}
If it's set and not empty then you can check for values that you need to check.

Answer 4

Try this:
<?php
$defaultvalue=0; // for example 0
$p=isset($_GET["p"]) ? $_GET["p"] : $defaultvalue;
if(($p != 'index') && ($p != 'reg') && ($p != 'login') && ($p != 'ad')):
?>