PHP glob（*）和foreach的最大数字basename

Question

I have this files path/0.php path/3.php path/2.php path/7.php

I want to display the files with foreach and I use this code:

$files = glob("system/$var/$var/$var/*/*.php");
foreach($files as $file) {
    //code here per $file.
    //for example:
    $basename = basename($file);
    echo str_replace(".php",null,$basename) . "<br>";
    include $file; //heres i need the path of the files.
}

But I want to sort this files by the largest number that is before the .php extension, like this:

Array
(
    [0] => path/7.php
    [1] => path/3.php
    [2] => path/2.php
    [3] => path/0.php
)

so?

update, rsort is not good for me, because the is a path/to/file, before the numbers, i want to use the path of the files.

Answer 1

You can sort the array being returned using usort:

$files = glob("system/$var/$var/$var/*/*.php"
usort($files, function ($b, $a) {
    return strcmp( substr($a, strrpos($a, '/') + 1), 
                   substr($b, strrpos($b, '/') + 1) ); 
});

Note that this will use string comparison, so 10.php will end up between 1.php and 2.php. If that's not what you want you have to use numeric comparison. i.e. using PHP 7's "spaceship" operator:

$files = glob("system/$var/$var/$var/*/*.php"
usort($files, function ($b, $a) {
    return ((int)substr($a, strrpos($a, '/') + 1)) 
       <=> ((int)substr($b, strrpos($b, '/') + 1)); 
});

---

Edit: I swapped $a and $b in the callback signatures to reverse the sorting order