Re: [PHP] Array help.

On Jul 30, 2010, at 2:36 PM, Paul Halliday wrote:

> I have a query that may not always return a result for a value, I need
> to reflect this with a "0". I am trying to overcome this by doing this
> (the keys are ID's):
> 
> while ($row = mysql_fetch_row($statusQuery)) {
> 
>        $cat = array(0=>0,1=>0,11=>0,12=>0,13=>0,14=>0,15=>0,16=>0,17=>0,19=>0);
> 
>        switch ($row[1]) {
>            case 0: $cat[0] = $row[0]; break;
>            case 1: $cat[1] = $row[0]; break;
>            case 11: $cat[11] = $row[0]; break;
>            case 12: $cat[12] = $row[0]; break;
>            case 13: $cat[13] = $row[0]; break;
>            case 14: $cat[14] = $row[0]; break;
>            case 15: $cat[15] = $row[0]; break;
>            case 16: $cat[16] = $row[0]; break;
>            case 17: $cat[17] = $row[0]; break;
>            case 19: $cat[19] = $row[0]; break;
>        }
> 
>        print_r($cat);
>    }
> 
> Which gives me this:
> 
> Array ( [0] => 15547 [1] => 0 [11] => 0 [12] => 0 [13] => 0 [14] => 0
> [15] => 0 [16] => 0 [17] => 0 [19] => 0 )
> Array ( [0] => 0 [1] => 0 [11] => 0 [12] => 0 [13] => 0 [14] => 0 [15]s
> => 30 [16] => 0 [17] => 0 [19] => 0 )
> 
> The query only return 2 rows:
> 
> 15 | 30
> 0 | 15547
> 
> What am I doing wrong? Is there a more elegant way to achieve what I want?
> 
> Thanks.
> 
> -- 
> Paul Halliday
> Ideation | Individualization | Learner | Achiever | Analytical
> http://www.pintumbler.org
> 
> -- 
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> 

Paul-

Why not say:

$cat = array();
if(isset($row[1])
{
    $cat[$row[1]] = $row[0];
}

print_r($cat);

Regards,

-Josh