为什么这个MySQL查询的结果相互成倍增加？

Question

SELECT user_id,
    SUM(COALESCE(point_points, 0)) AS total_points,
    SUM(
        CASE
            WHEN point_date > '$this_month'
            THEN point_points
            ELSE 0
        END)                AS month_points,
    COUNT(DISTINCT c_id)    AS num_comments,
    COUNT(DISTINCT rant_id) AS live_submissions
FROM users
    LEFT JOIN points
    ON  users.user_id = points.point_userid
    LEFT JOIN comments
    ON
        (
            c_userid = user_id
        )
    LEFT JOIN rants
    ON
        (
            rant_poster = user_id
        AND rant_status = 1
        )
WHERE user_id = $id
GROUP BY user_id

基本上live_submissions和num_comments可变显示正确的结果,而total_points与month_points显示器的产品month_points/total_points,live_submissions和num_comments.知道为什么会这样吗？

Answer 1

这被称为笛卡尔积.将表连接在一起时,默认结果是连接条件为true的每个行的排列.您使用JOIN条件来限制这些排列.

但是,由于您要连接多个表users,结果包括每个匹配表的每个排列.例如,每个匹配的行在每个匹配的行中points重复comments,并且每个匹配的行再次相乘,重复每个匹配的行rants.

您可以COUNT(DISTINCT c_id)像正在做的那样对此进行部分补偿,但这DISTINCT只是因为每个行有多行才有必要c_id.除非将其应用于唯一值,否则它不起作用.此补救措施不适用于SUM()表达式.

基本上,您试图在一个查询中进行太多计算.您需要将其拆分为单独的查询,以使其可靠.然后你也可以摆脱DISTINCT修饰符.

SELECT u.user_id, SUM(COALESCE(p.point_points, 0)) AS total_points, 
  SUM( CASE WHEN p.point_date > '$this_month' THEN p.point_points ELSE 0 END ) AS month_points
FROM users u LEFT JOIN points p
  ON u.user_id = p.point_userid 
WHERE u.user_id = $id
GROUP BY u.user_id;

SELECT user_id, COUNT(c.c_id) as num_comments, 
FROM users u LEFT JOIN comments c
  ON (c.c_userid = u.user_id)
WHERE u.user_id = $id
GROUP BY u.user_id;

SELECT u.user_id, COUNT(r.rant_id) as live_submissions
FROM users u LEFT JOIN rants r
  ON (r.rant_poster = u.user_id AND r.rant_status = 1)
WHERE u.user_id = $id
GROUP BY u.user_id;

您不应该尝试在单个查询中执行所有这三个操作.