in_array multiple values

04/15/2020 14:00:02

How do I check for multiple values, such as:

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


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

Verified Answer (196 Votes)

09/25/2011 04:51:50

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

Answer #2 (175 Votes)

06/15/2012 01:08:02

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

Answer #3 (13 Votes)

09/25/2011 04:51:06
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
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