为什么这个 MySQL 查询的结果会相互相乘？

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 and num_comments变量显示正确的结果，而total_points and month_points显示产品month_points/total_points, live_submissions and num_comments。知道为什么会发生这种情况吗？

这被称为笛卡尔积 http://en.wikipedia.org/wiki/Cartesian_product。当您将表连接在一起时，默认结果是每个行的排列其连接条件为真。你用JOIN限制这些排列的条件。

但是由于您要将多个表连接到users，结果包括每个匹配表的每个排列。例如，每个匹配行points中的每个匹配行重复comments，并且每一个都再次相乘，对每个匹配行重复rants.

您可以通过以下方式部分补偿COUNT(DISTINCT c_id)正如你所做的那样，但是DISTINCT是必要的，只是因为每个有多个行c_id。除非您将其应用于独特的值，否则它不会起作用。这个补救措施对以下情况不起作用SUM()表达式。

基本上，您试图在一个查询中执行太多计算。您需要将其拆分为单独的查询才能保证其可靠性。然后你就可以摆脱DISTINCT修饰符也是如此。

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完成所有这三个操作。