in_array multiple values

How do I check for multiple values, such as:

$arg = array('foo','bar');

if(in_array('foo','bar',$arg))

That's an example so you understand a bit better, I know it won't work.

Solution 1:

Intersect the targets with the haystack and make sure the intersection is precisely equal to the targets:

$haystack = array(...);

$target = array('foo', 'bar');

if(count(array_intersect($haystack, $target)) == count($target)){
    // all of $target is in $haystack
}

Note that you only need to verify the size of the resulting intersection is the same size as the array of target values to say that $haystack is a superset of $target.

To verify that at least one value in $target is also in $haystack, you can do this check:

 if(count(array_intersect($haystack, $target)) > 0){
     // at least one of $target is in $haystack
 }

Solution 2:

As a developer, you should probably start learning set operations (difference, union, intersection). You can imagine your array as one "set", and the keys you are searching for the other.

Check if ALL needles exist

function in_array_all($needles, $haystack) {
   return empty(array_diff($needles, $haystack));
}

echo in_array_all( [3,2,5], [5,8,3,1,2] ); // true, all 3, 2, 5 present
echo in_array_all( [3,2,5,9], [5,8,3,1,2] ); // false, since 9 is not present

Check if ANY of the needles exist

function in_array_any($needles, $haystack) {
   return !empty(array_intersect($needles, $haystack));
}

echo in_array_any( [3,9], [5,8,3,1,2] ); // true, since 3 is present
echo in_array_any( [4,9], [5,8,3,1,2] ); // false, neither 4 nor 9 is present

Solution 3:

if(in_array('foo',$arg) && in_array('bar',$arg)){
    //both of them are in $arg
}

if(in_array('foo',$arg) || in_array('bar',$arg)){
    //at least one of them are in $arg
}

Solution 4:

Going off of @Rok Kralj answer (best IMO) to check if any of needles exist in the haystack, you can use (bool) instead of !! which sometimes can be confusing during code review.

function in_array_any($needles, $haystack) {
   return (bool)array_intersect($needles, $haystack);
}

echo in_array_any( array(3,9), array(5,8,3,1,2) ); // true, since 3 is present
echo in_array_any( array(4,9), array(5,8,3,1,2) ); // false, neither 4 nor 9 is present

https://glot.io/snippets/f7dhw4kmju

Solution 5:

IMHO Mark Elliot's solution's best one for this problem. If you need to make more complex comparison operations between array elements AND you're on PHP 5.3, you might also think about something like the following:

<?php

// First Array To Compare
$a1 = array('foo','bar','c');

// Target Array
$b1 = array('foo','bar');


// Evaluation Function - we pass guard and target array
$b=true;
$test = function($x) use (&$b, $b1) {
        if (!in_array($x,$b1)) {
                $b=false;
        }
};


// Actual Test on array (can be repeated with others, but guard 
// needs to be initialized again, due to by reference assignment above)
array_walk($a1, $test);
var_dump($b);

This relies on a closure; comparison function can become much more powerful. Good luck!