獲取MySQL中一些行的平均值和單行的值

Question

我有一個有 4 列的表，如下所示：

ID - username - rating - solved

DB 用於拼圖網站。 ID是拼圖的#， username名字段是解決它的用戶，如果用戶已經解決，則solved的是 boolean。 在主頁上，我列出了謎題。 我得到了謎題的數據（來自另一個表），為登錄用戶解決的謎題值，但是當我嘗試獲得平均評分時，它平均一個值，即登錄用戶對謎題的評分，並忽略拼圖的任何其他評級。 這可以在一個查詢中完成，還是只在兩個查詢中完成，一個單獨的評級？ 如何？
哦，還有，如果沒有用戶登錄，這意味着我不必從評級表中獲取solved的值，評級是平均的，一切正常。

編輯：這是代碼。 $logged 是用戶是否登錄，所以前兩個查詢有效，后兩個查詢無效。

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：我嘗試了 2 個查詢，但是如果我嘗試 select solved的值在單獨的一個中，我無法通過AVG(rating)對其進行排序，我又回到了原來的問題。 有什么幫助嗎？

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

這將按當前登錄用戶的評分排序，單一個。

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