php - sort an array according to second given array

Question

$a = array('val1','val2',200, 179,230, 234, 242); 
$b = array(230, 234, 242, 179, 100);

So array $a should be sorted according to array $b and $resultArray should be ('val1','val2',200, 230, 234, 242, 179)

Answer 1

Your sorting requirements:

1. values not found in the sorting array come before values that ARE found.
2. then:
   1. sort found values by their position in the sorting array
   2. sort values not found in the sorting array normally/ascending

Flipping your order lookup array will permit improved efficiency while processing before key searching is faster than value searching in php.

Code: (Demo)

$array = ['val1', 'val2', 200, 179, 230, 234, 242]; 
$order = [230, 234, 242, 179, 100];
$keyedOrder = array_flip($order); // for efficiency

usort($array, function($a, $b) use ($keyedOrder) {
    return [$keyedOrder[$a] ?? -1, $a]
           <=>
           [$keyedOrder[$b] ?? -1, $b];
});
var_export($array);

Output:

array (
  0 => 'val1',
  1 => 'val2',
  2 => 200,
  3 => 230,
  4 => 234,
  5 => 242,
  6 => 179,
)

$keyedOrder[$variable] ?? -1 effectively means if the value is not found in the lookup, use -1 to position the item before the lowest value in the lookup array (0). If the value IS found as a key in the lookup, then use the integer value assigned to that key in the lookup.

---

From PHP7.4, arrow function syntax can be used to make the snippet more concise and avoid the use() declaration.

Demo

usort($array, fn($a, $b) => [$keyedOrder[$a] ?? -1, $a] <=> [$keyedOrder[$b] ?? -1, $b]);