mysql fetch array errors

[homers]

Errors:

Warning: mysql_fetch_array() expects parameter 1 to be resource, boolean given in C:\xampp\xampp\htdocs\index.php on line 188

 

188 rinda:

while ($row = mysql_fetch_array($results, MYSQL_ASSOC)) {

 

Viss kods:

<?php
$k = mysql_result(mysql_query("SELECT COUNT(*) as Num FROM komandas_turnira"),0);
$skaits = "8";
$s_rez = $skaits-$k;
$open = "o";
$results = mysql_query("SELECT id,nos,x FROM turniri WHERE status=".$open." ORDER BY id DESC");
while ($row = mysql_fetch_array($results, MYSQL_ASSOC)) {
printf("<img src='/style/icons/arrow_right.gif'> <a href=?p=cup&id=%s>%s Cup</a> [<font color=green>%s</font>/<font color=red>8</font>]<br/>", $row["id"], $row["nosaukums"], $e);
}
?>

Edited November 11, 2009 by homers

[briedis]

Kur tādu kodu var atrast :))

mysql_fetch_array($results, MYSQL_ASSOC)

 

manuprāt pietiek ar mysql_fetch_assoc($rez);

[homers]

To kodu var dabūt iekš php.net ..

Ieliekot tavu domāto kodu iznāk šads errors:

Warning: mysql_fetch_assoc() expects parameter 1 to be resource, boolean given in C:\xampp\xampp\htdocs\index.php on line 188

[xPtv45z]

Tad jau php.net arī varēji apskatīties, ko un kad atgriež mysql_query.

"For SELECT, SHOW, DESCRIBE, EXPLAIN and other statements returning resultset, mysql_query() returns a resource on success, or FALSE on error."

Edited November 11, 2009 by xPtv45z