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 $haystack
}
if(in_array('foo',$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 $arg
}
As a developer, you should probably start learning set operations (difference, union, intersection). You can imagine your array as one "set", and the keys you are searching for the other.
function in_array_all($needles, $haystack) {
return empty(array_diff($needles, $haystack));
}
echo in_array_all( [3,2,5], [5,8,3,1,2] ); // true, all 3, 2, 5 present
echo in_array_all( [3,2,5,9], [5,8,3,1,2] ); // false, since 9 is not present
function in_array_any($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
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!
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
if(empty(array_intersect([21,22,23,24], $check_with_this)) {
print "Not found even a single element";
} else {
print "Found an element";
}
array_intersect() returns an array containing all the values of array1 that are present in all the arguments. Note that keys are preserved.
Returns an array containing all of the values in array1 whose values exist in all of the parameters.
empty() — Determine whether a variable is empty
Returns FALSE if var exists and has a non-empty, non-zero value. Otherwise returns TRUE.
Source: Stackoverflow.com