Undefined variable problems

Neumy

I have a silly problem. I can't figure it out, maybe someone can help me.

This function is designed to make it easy for me to grab data out of my database. I just input what I want and let php do everything for me. My problem is when the sql did not find anything and returns a NULL value. When the program gets to:
if (!isset($data[1]))
$data = $data[0];
it calls up a "Notice: Undefined variable". I need this to go away and have tried all sorts of isset(), is_array(), ... with no luck. Can someone help me fix this?

Here's my function and function call:

$check = database_grab("data","table","this = '$that' AND what = '$why'","last_updated");

function database_grab($select,$table,$where,$order)
{
    $query = "SELECT $select FROM $table WHERE $where ORDER BY $order;";
    $result = mysql_query($query);
    $i=0;
    if (substr($select,0,9)=="DISTINCT ")
        $select = substr($select,9);
    $token = strtok($select,",");
    while($token !== false)
    {
        $field[] = $token;
        $token = strtok(",");
    }
    while($i<mysql_num_rows($result))
    {
        foreach($field as $val)
            $data[] = mysql_result($result,$i,$val);
        $i++;
    }
    if (!isset($data[1]))
        $data = $data[0];
    return $data;
} // function database_langGrab()

I'd appreciate any help you can give. Thanx!

troinfo

I would try adding something like this:


if (($value == "") || ($value == null)){

$value = 0;

} else {

// run your while loop

}

So instead of

 $data = $data[0];

use $data = 0;

I hope this helps some.

Some please correct me if I'm wrong or if I misread the script.

troinfo