有4列得到一些線路的平均單行在MySQL

Question

值我有一個表，如下所示：

ID - username - rating - solved 

的DB是一個謎網站。 ID是拼圖的＃，username字段是爲解決它的用戶，solved是一個布爾值，如果用戶已經解決了它。在主頁上，我列出了謎題。我從其他表中獲得了謎題的數據，這是解決登錄用戶謎題的值，但是當我嘗試獲得平均評級時，它將平均一個值，登錄用戶對謎題的評分，並忽略拼圖的任何其他評分。這可以在一個單一的查詢中完成，或者只有兩個單獨的查詢完成？怎麼樣？
另外，如果沒有用戶登錄，這意味着我不需要從收視率表中獲得solved值，收視率是平均值，一切正常。

編輯：這裏是代碼。 $ logged是如果用戶登錄的，所以前兩個查詢工作，最後兩個查詢不工作。

if ($logged) { 
    $query_recent = "SELECT id, owner, title, hidden, x, y, solved, points, AVG(rating) AS average FROM `users`, `cwords` LEFT JOIN `ratings` ON `cwords`.`id`=`ratings`.`cw_id` AND `ratings`.`owner_id` = '$username' WHERE `cwords`.`public` = 1 AND `owner` = `username` GROUP BY `id` ORDER BY `id` DESC LIMIT $from, 30"; 
    $query_top = "SELECT id, owner, title, hidden, x, y, solved, points, AVG(rating) AS average FROM `users`, `cwords` LEFT JOIN `ratings` ON `cwords`.`id`=`ratings`.`cw_id` AND `ratings`.`owner_id` = '$username' WHERE `cwords`.`public` = 1 AND `owner` = `username` GROUP BY `id` ORDER BY average DESC LIMIT $from, 30"; 
    } else { 
    $query_recent = "SELECT id, owner, title, hidden, x, y, points, AVG(rating) AS average FROM `users`, `cwords` LEFT JOIN `ratings` ON `cwords`.`id`=`ratings`.`cw_id` WHERE `cwords`.`public` = 1 AND `owner` = `username` GROUP BY `id` ORDER BY `id` DESC LIMIT $from, 30"; 
    $query_top = "SELECT id, owner, title, hidden, x, y, points, AVG(rating) AS average FROM `users`, `cwords` LEFT JOIN `ratings` ON `cwords`.`id`=`ratings`.`cw_id` WHERE `cwords`.`public` = 1 AND `owner` = `username` GROUP BY `id` ORDER BY average DESC LIMIT $from, 30"; 
    }; 

編輯2：我有2個查詢嘗試過，但如果我嘗試在一個單獨的一個選擇solved值我不能由AVG(rating)爲了它，我回到原來的問題。任何幫助？

$query_top_solved = "SELECT id, solved FROM cwords LEFT JOIN ratings ON `cwords`.`id`=`ratings`.`cw_id` AND `ratings`.`owner_id` = '$username' WHERE `cwords`.`public` = 1 GROUP BY cw_id ORDER BY AVG(rating) DESC LIMIT $from, 30"; 

這可以通過當前登錄的用戶評分（單一評分）對其進行排序。

Answer 1

您需要兩個單獨的查詢。一個獲取用戶的分數，一個獲取總體平均分數。

SELECT rating, solved 
FROM ... 
WHERE username=$userID AND ID=$puzzleID 

和

SELECT AVG(rating), AVG(solved) 
FROM ... 
WHERE ID=$puzzleID