Question

我有一个包含4列的表，如下所示：

ID - username - rating - solved

数据库适用于拼图网站。 ID是拼图的＃，username字段是解决它的用户，solved是布尔值，如果用户已经解决了它。在主页面上，我列出了谜题。我得到了谜题的数据（来自另一张表），登录用户的谜题的解决值，但是当我尝试获得平均评级时，它平均一个值，即登录用户对谜题的评级，并忽略了拼图的任何其他评级。这可以在一个查询中完成，还是只有两个，一个单独的查询？怎么样？
哦，如果没有用户登录，这意味着我不必从评级表中获取solved值，评级是平均的，一切正常。

编辑：这是代码。 $ logged是用户登录的，因此前两个查询有效，而后两个查询没有。

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次查询，但是如果我尝试在另一个中选择solved值，我无法通过AVG(rating)订购，我又回到原来的问题。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";

按当前登录用户的评分单一订购。

Answer 1

您需要两个单独的查询。一个用于获取用户的分数，另一个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

获取MySQL中某些行的平均值和单行的值

1 个答案: