Просмотр поста

Ошибка 1:
Warning: mysql_result(): supplied argument is not a valid MySQL result resource in ....


Запрос с ошибкой:
$u = mysql_result(mysql_query("SELECT COUNT(*) FROM `m_game` where `id_team1`= '" . $fid ."' OR `id_team2`= '" . $fid ."' ;"),0);