Why is array_search not finding, unless array is sorted?

Question

I'm having trouble understanding the result of array_search in the following example (and did not find any existing questions discussing this):

<?php

    $values = array(
        15,
        12,
        "15",
        34,
        15 => 25,
        "xx" => 15
    );
    echo "PHP-Version is " . phpversion();
    echo "<h1>Array:</h1><pre>";var_dump($values);echo "</pre>";    

    // sort($values); // remove comment and 15 will be found in ALL cases!

    $key = array_search("15",$values);
        show_result('(a) Searching "15"');

    $key = array_search("15",$values,true);
        show_result('(b) Searching "15",true');

    $key = array_search(15,$values);
        show_result('(c) Searching 15');

    $key = array_search(15,$values,false);
        show_result('(d) Searching 15,false');

    $key = array_search(15,$values,true);
        show_result('(e) Searching 15,true');



function show_result($tit) {    
    global $key,$values;
    echo "<h2>$tit</h2>";

    if (!$key) { 
        echo "Not found";
        } else {
            echo "Found key $key - " . gettype($values[$key]);
    }
}
?>

Only search (b) - the strict string-search finds the value, numeric search does not find it. All searches do find it when the array is sorted - but the doc does not mention such a requirement at all. Can someone explain this behaviour?

For once the OP already read the manual. Too bad you didn't read it in its entirety. ;-P — deceze, Aug 06 '13 at 08:29

score 4 · Accepted Answer · answered Aug 06 '13 at 08:25

4

The value 15 is at key 0. array_search returns this 0. 0 evaluates to false. Your check if (!$key) therefore fails for the key 0. You have to check for strict === false. There's a giant red warning in the manual explaining this.

answered Aug 06 '13 at 08:25

deceze

510,633
85
743
889

Thank you - I was so focused on the array that I did not pay enough attention to the type of the result. – MBaas Aug 06 '13 at 08:34

score 1 · Answer 2 · answered Aug 06 '13 at 08:25

1

I think this is because the return value of array_search is a mixed value. When you perform the first search it will find the value at position 0 and therefor return the keys index which in turn will be evaluated to false.

answered Aug 06 '13 at 08:25

Cyclonecode

29,115
11
72
93

score 1 · Answer 3 · answered Aug 06 '13 at 08:32

Here's the total explaination:

In all your cases, $key equals to 0, and since you are doing a simple test like if($key) 0 evaluates to false making you think that the search was not succesful. Changing it to if($key===false) will work without any problem.

The reason why it works without any problem when you sort the array is because the value 12 is inside so it takes the key 0 and the 15 that was at key 0 is now at key 1 and 1 evaluates to true.

Why is array_search not finding, unless array is sorted?

3 Answers3