Is there a way to simplify this case statement?

Question

I have this PHP case statement

switch ($parts[count($parts) - 1]) {
    case 'restaurant_pos':
        include($_SERVER['DOCUMENT_ROOT'] . '/pages/restaurant_pos.php');
        break;
    case 'retail_pos':
    include($_SERVER['DOCUMENT_ROOT'] . '/pages/retail_pos.php');
        break;  
    .....

}

Which works great but I have many many files (like 190) and I would love to know if there is a way to make this case statement many work with anything so I dont have to do 190 case conditions. I was thinking I can use the condtion in the case and maybe see if that file is present and if so then display and if not then maybe a 404 page but i was not sure a good way to do this...any ideas would help alot

Answer 1

You can predefine file names in an array and then use in_array in order to check name's existence:

$files = array('restaurant_pos', 'retail_pos', ......);
$file = $parts[count($parts) - 1];
if (in_array($file, $files)) {
    include($_SERVER['DOCUMENT_ROOT'] . "/pages/$file.php");
}

Answer 2

If it's not user input, you can do it like

$include = $parts[count($parts) - 1];
if ($include) {
    if (file_exists($_SERVER['DOCUMENT_ROOT'] . '/pages/'.$include.'.php')){
          include $_SERVER['DOCUMENT_ROOT'] . '/pages/'.$include.'.php';
    }
}

repeating, don't do this if $include is being filled from user's input !