PHP: Check if an array contains all array values from another array

Question

$all = array
(
    0 => 307,
    1 => 157,
    2 => 234,
    3 => 200,
    4 => 322,
    5 => 324
);
$search_this = array
(
    0 => 200,
    1 => 234
);

I would like to find out if $all contains all $search_this values and return true or false. Any ideas please?

Answer 1

Look at array_intersect().

$containsSearch = count(array_intersect($search_this, $all)) === count($search_this);

Answer 2

The previous answers are all doing more work than they need to. Just use array_diff. This is the simplest way to do it:

$containsAllValues = !array_diff($search_this, $all);

That's all you have to do.

Answer 3

A bit shorter with array_diff

$musthave = array('a','b');
$test1 = array('a','b','c');
$test2 = array('a','c');

$containsAllNeeded = 0 == count(array_diff($musthave, $test1));

// this is TRUE

$containsAllNeeded = 0 == count(array_diff($musthave, $test2));

// this is FALSE

Answer 4

I think you're looking for the intersect function

array array_intersect ( array $array1 , array $array2 [, array $ ... ] )

array_intersect() returns an array containing all values of array1 that are present in all the arguments. Note that keys are preserved.

http://www.php.net/manual/en/function.array-intersect.php

Answer 5

How about this:

function array_keys_exist($searchForKeys = array(), $searchableArray) {
    $searchableArrayKeys = array_keys($searchableArray);

    return count(array_intersect($searchForKeys, $searchableArrayKeys)) == count($searchForKeys); 
}