Fetch array with mysql_fetch_assoc

Question

$query = mysql_query("SELECT Name from lottery");
// Make All Lotteries Into An Array
$queryarray = mysql_fetch_array($query);
// For Each Lottery, Store the Name of It in $lottery
foreach ($queryarray as $lottery) {
    // Select IDs of Tickets in Current Selected Lottery ($lottery)
    $ticketquery = mysql_query("SELECT id FROM tickets WHERE lottery='$lottery'");
    // Create an Array from the IDS
    $ticketarray = mysql_fetch_assoc($ticketquery);
    // Select A Random ID, This Is Our Winner Ticket
    $winner = $ticketarray[array_rand($ticketarray)];
}

When I run this, however, I get this error:

"Warning: array_rand() [function.array-rand]: First argument has to be an array in public_html/php/lotterypick.php on line 25"

I have tried to replace mysql_fetch_assoc with mysql_fetch_array and that hasn't helped much either. Just given me the same error. Any ideas? Thanks.

Line 25: $winner = $ticketarray[array_rand($ticketarray)];

Put var_dump($queryarray);die; before the loop, and see what it contains... it does not contain what you think it contains. — Sverri M. Olsen
– Sverri M. Olsen, Commented Mar 28, 2015 at 19:30
Oh wow. Yeah, that is definitely not what I thought. How would I fix that? @SverriM.Olsen — TonyStark
– TonyStark, Commented Mar 28, 2015 at 19:33
But I don't understand how combing rows from the tables would solve this issue? Or would it? @eggyal — TonyStark
– TonyStark, Commented Mar 28, 2015 at 19:36
@TonyStark: Or you could just ask MySQL to select a random ticket from each lottery, as one query. — eggyal
– eggyal, Commented Mar 28, 2015 at 19:46

bayblade567 · Accepted Answer · 2015-03-28 20:51:20Z

1

Writing the code as:

while($lottery = mysql_fetch_array($query)){...}

would solve the problem.

answered Mar 28, 2015 at 20:51

bayblade567

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