Question

我有这些表：users，comments，ratings和items

我想知道是否可以编写基本上执行此操作的SQL查询：

user_id在每个表中。我想要一个SQL查询来计算每个表中的每个出现次数（当然除了users）。但是，我想要一些桌子比其他桌子更重。然后我想计算一个“得分”。

以下是一个例子：

user_id 5发生... 物品2次; 5次评论; 11次评级。

我想要一个总计如下的公式/点数系统：

项目2 x 5 = 10; 评论5 x 1 = 5; 评级11 x .5 = 5.5

总计21.5

这就是我到目前为止......

SELECT u.users
     COUNT(*) r.user_id
     COUNT(*) c.user_id
     COUNT(*) i.user_id
FROM users as u
JOIN COMMENTS as c
     ON u.user_id = c_user_id
JOIN RATINGS as r
     ON r.user_id = u.user_id
JOIN ITEMS as i
     i.user_id = u.user_id
WHERE
    ????
GROUP BY u.user_id
ORDER by total DESC

我不知道如何做数学公式部分（如果可能的话）。或者如何计算总数。

基于John Woo的答案的最终守则！

$sql = mysql_query("
        SELECT  u.username,
    (a.totalCount * 5) +
    (b.totalCount) +
    (c.totalCount * .2) totalScore
    FROM users u 
    LEFT JOIN
        (
            SELECT user_id, COUNT(user_id) totalCount
            FROM items 
            GROUP BY user_id
        ) a ON a.user_id= u.user_id
    LEFT JOIN
        (
            SELECT user_id, COUNT(user_id) totalCount
            FROM comments
            GROUP BY user_id
        ) b ON b.user_id= u.user_id
    LEFT JOIN
        (
            SELECT user_id, COUNT(user_id) totalCount
            FROM ratings
            GROUP BY user_id
        ) c ON c.user_id = u.user_id
    ORDER BY totalScore DESC LIMIT 10;");

Answer 1

也许这可以帮到你，

SELECT  u.user_ID,
        (a.totalCount * 5) +
        (b.totalCount) +
        (c.totalCount * .2) totalScore
FROM    users u LEFT JOIN
            (
                SELECT user_ID, COUNT(user_ID) totalCount
                FROM items 
                GROUP BY user_ID
            ) a ON a.user_ID = u.user_ID
        LEFT JOIN
            (
                SELECT user_ID, COUNT(user_ID) totalCount
                FROM comments
                GROUP BY user_ID
            ) b ON b.user_ID = u.user_ID
        LEFT JOIN
            (
                SELECT user_ID, COUNT(user_ID) totalCount
                FROM ratings
                GROUP BY user_ID
            ) c ON c.user_ID = u.user_ID
ORDER BY totalScore DESC

但基于上面的yur查询，这可能也有效

SELECT  u.users
        (COUNT(*) * .5) +
        COUNT(*) +
        (COUNT(*) * 2) totalcore
FROM users as u
        LEFT JOIN COMMENTS as c
            ON u.user_id = c_user_id
        LEFT JOIN RATINGS as r
            ON r.user_id = u.user_id
        LEFT JOIN ITEMS as i
            ON i.user_id = u.user_id
GROUP BY u.user_id
ORDER by totalcore DESC

唯一的区别是使用LEFT JOIN。在这种情况下，您不会使用INNER JOIN，因为有可能无法保证每个表都存在user_id。

希望这是有道理的

enter image description here

Answer 2

这是另一种方法：

SELECT
  u.user_id,
  SUM(s.weight) AS totalScore
FROM users u
  LEFT JOIN (
    SELECT user_id, 5.0 AS weight
    FROM items
    UNION ALL
    SELECT user_id, 1.0
    FROM comments
    UNION ALL
    SELECT user_id, 0.5
    FROM ratings
  ) s
  ON u.user_id = s.user_id
GROUP BY
  u.user_id

即。对于每个表中每个用户的每次出现，都会产生一个具有特定权重的行。然后将UNIONed权重集连接到users表，以便进行后续分组和聚合。

在SQL中编写数学公式

2 个答案: