Conditional statement not working

[[deleted]]

This should be so easy, but it is not working.

What should happen is that if $p=home in the url then it displays home.php and if $p does not equal home then it displays page.php which is a template.

What am I missing?

<?php
// display page

if (@$p == $_GET['home']){
	include ('pages/home.php');
} elseif (@$p != $_GET['home']){
	include ('pages/page.php');
}else{
		include ('pages/home.php');
}
?>

[LordShryku]

} elseif (@$p != $_GET['home']){

So uhm...if $_GET['home'] doesn't exist, this will fail, which will then send you to your else {

[zcketcher]

How I would do it:

$path = "pages/home.php";
if(@$p != $_GET["home"]) $path = "pages/page.php";
include("$path");

No need for any if/else...just one if.

[greg252]

Switch will make the code a bit easier to read.

[BSTRhino]

If I understand your explanation correctly, I think the code is a little muddled up.

It looks like what you want is for http://www.somewhere.com/something.php?p=home to go to pages/home.php is that right?

In that case, I would do your code like this.

<?php

if($_GET['p'] == 'page') {
    include 'pages/page.php';
}
else{
    include 'pages/home.php';
}

?>

I think what was causing the code to function improperly was the fact that $_GET['home'] == $p means:
http://www.somewhere.com/something.php?home=$p

whereas you had wanted this:
http://www.somewhere.com/something.php?$p=home

Just remember that the 'p' bit in $_GET['p'] refers to the name of the parameter p, what is returned is the value of this parameter.