PHP: array_search returning null

Question

I have a regular array that contains numbers from 1 to 14 that is generated by retrieving INTs from a SQL result column.

Using another query I'm returning another INT.

I'm searching for this INT in my array.

$key = array_search("$roomNb", $freeRooms);

However when I try to echo this key nothing appears. It is empty.

If I echo $freeRooms using a loop I get:

If I echo $roomNb I get

So I can't understand why I'm not getting anything in return. I should be expecting a key of 4 no? What could be causing this?

Don't just use echo - use var_dump($roomNb, $freeRooms). I'm willing to bet there's an extra space or other character in your needle or haystack. In general, var_dump can help you fix all your code problems :-) — cmbuckley
– cmbuckley, Commented Aug 30, 2013 at 22:26
@Itay quotes make no difference — array_search is non-strict by default. (OK, technically not true, if the array contained string("5 ") and $roomNb was int(5), then this would make a difference…) — cmbuckley
– cmbuckley, Commented Aug 30, 2013 at 22:28
I just replicated the scenario in my terminal and it worked fine, maybe try casting (int) when you're storing the values in the array? — Sterling Archer
– Sterling Archer, Commented Aug 30, 2013 at 22:32
@cbuckley - I think even in that case it's OK. $a = ['4', '5 ', '6']; echo array_search(5, $a); that echoes 1, since PHP's string->int conversion looks for the first non-whitespace character and then checks if that is the start of a sequence of digits before the next non-digit character matching the search input. It would even work if the value in the array were '5g' or ' 5 6' (space between the 5 and 6). — user2136840
– user2136840, Commented Aug 30, 2013 at 22:43
@ChicagoRedSox yes, my point was that the quotes would make a difference in that scenario, as array_search('5', ['5 ']) would return false! There was a deleted comment that said removing the quotes around $roomNb would fix it. — cmbuckley
– cmbuckley, Commented Aug 31, 2013 at 10:20

Matt Urtnowski · Accepted Answer · 2013-08-30 23:24:12Z

1

Most likely $key is false which when sent to echo will appear as nothing. Instead you should use var_dump to determine the value of $key

False is returned from array_search when the needle is not found

Null is only returned when the parameters you pass to array_search are invalid

http://php.net/manual/en/function.array-search.php

Test this code here using http://writecodeonline.com/php/:

echo false;
echo null;
var_dump(false);
var_dump(null);

$code = array(1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14);
$roomNb = 5;

var_dump(array_search("$roomNb", $code));

Should work just fine. echo will output nothing. var_dump will output stuff. And the key is found at 4.

edited Aug 30, 2013 at 23:24

answered Aug 30, 2013 at 23:07

Matt Urtnowski

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DigitalJedi805 · Accepted Answer · 2013-08-30 22:42:34Z

0

Assuming that your array will always contain contiguous numbers, and you can verify that your second SQL call returns one of those numbers, I would say something like

$value = $freeRooms[$roomNb]

I'm personally unfamiliar with how array_search executes, however if you wanted to write your own, I assume it would look something like this

function IndexOf($searchValue,$array)
{
    $i = 0;
    for( $i = 0; $i < count($array); $i++ ) 
        if( $array[$i] == $searchValue )
            return $i;
}

answered Aug 30, 2013 at 22:42

DigitalJedi805

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1 Comment

user2136840 Over a year ago

The array doesn't contain contiguous numbers: it starts with 1, not 0, and it skips 13.

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