Question

I have a simple if statement that works on my local machine but when i uploaded it tom ypaid site i got an error on my home page.

Error:

Warning: include(./pages/.php) [function.include]: failed to open stream: No such file or directory in /home/a5410474/public_html/index.php on line 33
Warning: include() [function.include]: Failed opening './pages/.php' for inclusion (include_path='.:/usr/lib/php:/usr/local/lib/php') in /home/a5410474/public_html/index.php on line 33

The Code:

<?php
    if (isset($_GET['p']) && $_GET['p'] != NULL || $_GET['p'] != '') {
        $p = mysql_real_escape_string($_GET['p']);
    }elseif ($_GET['p'] == '' || $_GET['p'] == NULL){
        $p = 'home';
    }
include("./pages/".$p.".php");
?>

Answer 1

A much less verbose way of writing this:

$p = empty($_GET['p']) ? 'home' : $_GET['p'];
include("./pages/{$p}.php");

Some other notes:

1. You should not be using mysql_real_escape_string on a variable you’re going to pass to include. That function is for preparing data for insertion into a SQL query.

2. You should not include a file based on a variable passed through the query string, or from any kind of user input. Someone can use that to read system files on your server then take control of the whole computer.