PHP code to exclude index.php using glob

Question

Problem

I am trying to display a random page from a file called ../health/ In this file there is a index.php file and 118 other files named php f

Answer 1

The first thing that came to mind is array_filter(), actually it was preg_grep(), but that doesn't matter:

$health = array_filter(glob("../health/*.php"), function($v) {
    return false === strpos($v, 'index.php');
});

With preg_grep() using PREG_GREP_INVERT to exclude the pattern:

$health = preg_grep('/index\.php$/', glob('../health/*.php'), PREG_GREP_INVERT);

It avoids having to use a callback though practically it will likely have the same performance

Update

The full code that should work for your particular case:

$health = preg_grep('/index\.php$/', glob('../health/*.php'), PREG_GREP_INVERT);
$whathealth = $health[mt_rand(0, count($health) -1)];
include ($whathealth);

Answer 2

To compliment Jack's answer, with preg_grep() you can also do:

$files = array_values( preg_grep( '/^((?!index.php).)*$/', glob("*.php") ) );

This will return an array with all files that do NOT match index.php directly. This is how you could invert the search for index.php without the PREG_GREP_INVERT flag.