php - MySQL中获取部分行的平均值和单行的值

Question

我有一个包含 4 列的表格，如下所示:

ID - username - rating - solved

数据库用于拼图网站。 ID 是谜题的编号，username 字段用于解决该谜题的用户，solved 是一个 bool 值(如果用户已解决)它。在主页上，我列出了谜题。我获取谜题的数据(从另一张表)，即登录用户的谜题的已解决值，但是当我尝试获取平均评分时，它会平均一个值，即登录用户对谜题的评分，并忽略该谜题的任何其他评级。这可以在一个查询中完成，还是只能在两个查询中完成，一个单独的评级查询？怎么办？
哦还有，如果没有用户登录，这意味着我不必从评级表中获取已解决值，评级会被平均，一切都会正常。

编辑:这是代码。 $logged 表示用户是否已登录，因此前两个查询有效，而后两个查询无效。

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";
  };

编辑2:我尝试了2个查询，但如果我尝试在单独的查询中选择已解决值，我无法按AVG(评级)对其进行排序，我又回到了原来的问题。有什么帮助吗？

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

这会根据当前登录用户的评分(单个评分)进行排序。

Answer 1

您需要两个单独的查询。一种用于获取用户的分数，另一种用于获取总体平均分数。

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

和

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