php in_array - unexpected behavior

Question

Not sure why this is happening, but my script doesn't seem to be able to return true for in_array more than once...

$saved = '15,22';    
$set = explode(",",$saved); //results in Array ( [0] => 15 [1] => 22 )

Then, I query the database:

$result = pg_query("SELECT did,vid,iid,value FROM demographicValues");
if(pg_num_rows($result) > 0) {
  while($r = pg_fetch_array($result)) {
     $demo[$r['did']][$r['vid']]['value'] = $r['value'];
     if(in_array($r['vid'], $set)) {
       $demo[$r['did']][$r['vid']]['status'] = 1;
     }
  }
} else...

If I print_r $demo, you can see that the vid 22 is in there, so, I'm not understanding why the status isn't being set accordingly?

    Array
    (
        [Mant] => Array
            (
                [15] => Array
                    (
                        [value] => Proper
                        [checked] => 1
                    )

                [16] => Array
                    (
                        [value] => Parish
                    )
        [Comp] => Array
            (
                [22] => Array
                    (
                        [value] => 65 - 70
                    )

                [23] => Array
                    (
                        [value] => 35 - 50
                    )
        )

)

Note that I also tried array_intersect and array_flip on $set, then isset...

There may be some spaces or other invisible characters messing up the code. Can you use var_dump rather than print_r? It'll put quotes around the strings and tell us the number of variables. — Samir Talwar
– Samir Talwar, Commented Mar 30, 2010 at 13:08
if you add var_dump($r['vid'], $set) immediately before 'in_array' line - what does it say? — user187291
– user187291, Commented Mar 30, 2010 at 13:10
Thank you so much... Yes, the $set did have some unnecessary spaces, and using trim got it to work... Thanks! — n00b0101
– n00b0101, Commented Mar 30, 2010 at 13:11
normally, I'd recommend using an associative array over in_array for performance reasons($set = array_flip(explode(',', $saved)); ... if (isset($set[$r['vid']]))), but $set is so short it won't make much of a difference; in_array might even be faster for just a few elements. Another approach is to set the status in the query: SELECT did,vid,iid,value, vid IN ($saved) AS status FROM demographicValues; — outis
– outis, Commented Mar 30, 2010 at 13:30
@Sterofrog and @n00b0101: suggest reposting comment as answer and select as correct answer so the question no longer shows up on unanswered. — JYelton
– JYelton, Commented Apr 16, 2010 at 20:52

Kru · Accepted Answer · 2010-04-28 20:29:42Z

2

By default in_array also checks the types. See the parameter strict at http://php.net/manual/en/function.in-array.php.

In your code $r['vid'] is an integer and the exploded string is still a string. So just use the same types or use:

if(in_array($r['vid'], $set, false)) ...

answered Apr 28, 2010 at 20:29

Kru

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1 Comment

knittl Over a year ago

by default php will not check the types

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