How can I use in_array if the needle is an array?

℡╲_俬逩灬. 提交于 2019-11-27 12:33:56

Use array_diff():

$arr1 = array(1,2,3);
$arr2 = array(1,2,3,4,5,6,7);
$arr3 = array_diff($arr1, $arr2);
if (count($arr3) == 0) {
  // all of $arr1 is in $arr2
}

You can use array_intersect or array_diff:

$arr1 = array(1,2,3);
$arr2 = array(1,2,3,4,5,6,7);

if ( $arr1 == array_intersect($arr1, $arr2) ) {
    // All elements of arr1 are in arr2
}

However, if you don't need to use the result of the intersection (which seems to be your case), it is more space and time efficient to use array_diff:

$arr1 = array(1,2,3);
$arr2 = array(1,2,3,4,5,6,7);
$diff = array_diff($arr1, $arr2);

if ( empty($diff) ) {
    // All elements of arr1 are in arr2
}

You can try use the array_diff() function to find the difference between the two arrays, this might help you. I think to clarify you mean, all the values in the first array must be in the second array, but not the other way around.

In my particular case I needed to check if a pair of ids was processed before or not. So simple array_diff() did not work for me.

Instead I generated keys from ids sorted alphabetically and used them with in_array:

<?php
$pairs = array();
// ...
$pair = array($currentId, $id);
sort($pair);
$pair = implode('-', $pair);
if (in_array($pair, $pairs)) {
    continue;
}
$pairs[$pair] = $pair;

This is probably not an optimum solution at all but I just needed it for a dirty script to be executed once.

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