我試着按前面的鏈接建議組合,但我仍然收到錯誤。我相當新的PHP,所以這就是爲什麼我有兩個querys。想改變顯示的統計順序
警告:mysql_fetch_assoc():提供的參數不是在/home/vhockey/public_html/vhatest/connect.php一個有效的MySQL結果資源上線88
該表是 「season12」 和表是「p」
這裏是除了服務器信息我connect.php文件...
function index_team_stats($subconference) {
$return = array();
$query = "SELECT id, teamname, teamnameseason, teamabr
FROM teams
WHERE subconference = '" . $subconference . "'
ORDER BY teamnameseason";
$teams = result_array($query);
foreach ($teams as $team)
{
$query = "SELECT gp, w, l, ol, p
FROM season12
WHERE team = '" . $team['teamnameseason'] . "'
ORDER BY p DESC
LIMIT 0,20'; ";
$results = result_array($query);
if ($results)
{
$results[0]['team'] = str_replace($team['teamnameseason'], '', $team['teamname']);
$results[0]['teamabr'] = $team['teamabr'];
$results[0]['teamid'] = $team['id'];
$return[] = $results[0];
}
}
return $return;
}
function get_team_name($teamnameseason) {
$query = "SELECT teamname FROM teams WHERE teamnameseason = '" . $teamnameseason . "'";
$row = mysql_fetch_row(mysql_query($query));
return str_replace($teamnameseason, '', $row[0]);
}
function result_array($query) {
$results = mysql_query($query) or die("error on: " . $query . " saying: " . mysql_error());
$return = array();
while ($row = mysql_fetch_assoc($results)) {
$return[] = $row;
}
return $return;
}
這裏是信息的圖像![由團隊PTS排序的所有信息從最高到最低
應該表現出企鵝,傳單,島民,流浪者,紅魔..
我會用一個加盟者2個SQL查詢合二爲一。 – 2012-02-28 16:30:44
是'result_array'用戶函數嗎?或者你混淆了'mysql_fetch_array' – 2012-02-28 16:32:44