使用array_search（）返回与搜索字符串不匹配的第一个元素的键

Question

If there is a way by which array_search() would return key of its first conflict like if I run

$key = array_search(40489, array_column($userdb, 'uid'));

on

Array
(

    (1) => Array
        (
            (uid) => '5465',
            (name) => 'Stefanie Mcmohn',
            (pic_square) => 'urlof100'
        ),

    (2) => Array
        (
            (uid) => '40489',
            (name) => 'Michael',
            (pic_square) => 'urlof40489'
        )
);

it would ideally return

2

but I want it to return

1

i.e the first element that did not have the 'uid' = 40489 &

If its not possible with array_search() is there any other way to do it with loops? I tried array_filter() but can't get it to work.

Answer 1

As said in array_search description:

Searches the array for a given value and returns the first corresponding key if successful

Returns the key for needle if it is found in the array, FALSE otherwise.

So, you cannot use array_search to search something that not equals what you need. Instead write you own function, for example:

$array = [];    // your array
foreach ($array as $key => $value) {
    if ($value['uid'] != '40489') {
        echo 'Key: ', $key;
        // use `break` to stop iterating over 
        // array as you already found what you need
        break;
    }
}