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?

Possible duplicate of [Checking to see if one array's elements are in another array in PHP](http://stackoverflow.com/questions/523796/checking-to-see-if-one-arrays-elements-are-in-another-array-in-php) — Vishal Kumar Sahu, Apr 19 '17 at 10:01
@VishalKumarSahu Not quite a duplicate: Your given link has to do with checking if ANY elements are contained in another array, not if ALL elements are contained in another. — Stefan, Jun 20 '17 at 15:15

score 295 · Answer 1 · edited Mar 03 '16 at 15:10

295

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.

edited Mar 03 '16 at 15:10

Zanshin13

821
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answered Mar 26 '14 at 03:50

orrd

8,689
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5

Thanks for the aha moment; I came in thinking `array_intersect`. – Derek Illchuk Jul 04 '14 at 22:24
21

this should be the answer – Matthew Oct 17 '15 at 07:28
2

Yes it's such better solution – Oleksandr Diudiun Dec 22 '15 at 09:57
1

Better solution than intersect ! – TheBosti Feb 26 '17 at 21:01
1

I came across this page and used the above solution because it accounted for duplicates – Rayzor Feb 28 '18 at 01:36
1

Indeed, this should be the accepted answer – Pierre B. Dec 29 '20 at 10:44
1

What about of using `empty(array_diff($search_this, $all));`? – Miguel May 06 '21 at 08:34
! operator on array will always return true. Use @Miguel 's suggestion: empty(array_diff($search_this, $all)); – cquezel Jan 14 '22 at 19:56
The original answer works because an empty array evaluates as false in PHP. Using `empty()` also works, but the `!` operator does work fine. – orrd Jan 17 '22 at 01:35
@Miguel an empty array in this context is treated as falsey, there is no need to use `empty`. – zanderwar Apr 08 '22 at 13:17

score 190 · Accepted Answer · edited Mar 31 '22 at 14:16

190

Look at array_intersect().

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

Or for associative array, look at array_intersect_assoc().

Or for recursive compare of sub-arrays, try:

<?php

namespace App\helpers;

class Common {
    /**
     * Recursively checks whether $actual parameter includes $expected.
     *
     * @param array|mixed $expected Expected value pattern.
     * @param array|mixed $actual Real value.
     * @return bool
     */
    public static function intersectsDeep(&$expected, &$actual): bool {
        if (is_array($expected) && is_array($actual)) {
            foreach ($expected as $key => $value) {
                if (!static::intersectsDeep($value, $actual[$key])) {
                    return false;
                }
            }
            return true;
        } elseif (is_array($expected) || is_array($actual)) {
            return false;
        }
        return (string) $expected == (string) $actual;
    }
}

edited Mar 31 '22 at 14:16

Top-Master

5,262
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answered Mar 11 '12 at 14:21

jasonbar

12,945
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13

You know you can omit both `count()` calls? – Wrikken Aug 15 '13 at 16:01
1

@Wrikken Can't the values get reordered during `array_intersect()`? I mean, `['a', 'b'] != ['b', 'a']`. – sbichenko Oct 16 '13 at 18:57
1

@exizt: `array_intersect()` does not alter the input arrays, so `$search_this` & `$all` are safe (it just returns an output). The function signature is `array array_intersect ( array $array1 , array $array2 [, array $... ] )` (safe). If it would/could alter them, it would be `array array_intersect ( array &$array1 , array &$array2 [, array &$... ] )` (possible altering of input arguments). Also, the keys of `$search_this` are _preserve_, and the order of the first array is kept. So, both key/value pairs, as their order, match. – Wrikken Oct 16 '13 at 21:03
4

And even then: [array comparison](http://php.net/manual/en/language.operators.array.php): _"`==` TRUE if $a and $b have the same key/value pairs."_, so the order doesn't even matter (use `===` for that) – Wrikken Oct 16 '13 at 21:24
2

This answer assumes that the $all array only contains unique values. If this is not the case, one may use the array_unique function on the $all array in the array_intersects function. – Relequestual Dec 12 '13 at 21:49
5

array_diff solution @orrd gave is much cleaner – Joseph Connolly Apr 05 '18 at 18:29

score 14 · Answer 3 · answered Mar 10 '14 at 17:48

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

score 4 · Answer 4 · edited Jul 12 '16 at 07:18

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

edited Jul 12 '16 at 07:18

shA.t

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answered Mar 11 '12 at 14:19

James Kyburz

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score 0 · Answer 5 · edited Jul 26 '20 at 13:17

0

How about this:

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

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

edited Jul 26 '20 at 13:17

MaartenDev

5,093
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answered Nov 18 '17 at 14:03

K-Alex

350
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PHP: Check if an array contains all array values from another array

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