来自GROUP_BY的两个LEFT JOIN的GROUP_CONCAT的奇怪重复行为

Question

这是我所有表格的结构和查询(请关注最后一个查询,如下所示).正如你在小提琴中看到的,这是当前的输出:

+---------+-----------+-------+------------+--------------+
| user_id | user_name | score | reputation | top_two_tags |
+---------+-----------+-------+------------+--------------+
| 1       | Jack      | 0     | 18         | css,mysql    |
| 4       | James     | 1     | 5          | html         |
| 2       | Peter     | 0     | 0          | null         |
| 3       | Ali       | 0     | 0          | null         |
+---------+-----------+-------+------------+--------------+

这是正确的,一切都很好.

---

现在我还有一个名为"category"的存在.每个帖子只能有一个类别.而且我还希望为每个用户获得前两个类别.而这里是我的新的查询.正如您在结果中看到的,发生了一些重复:

+---------+-----------+-------+------------+--------------+------------------------+
| user_id | user_name | score | reputation | top_two_tags |   top_two_categories   |
+---------+-----------+-------+------------+--------------+------------------------+
| 1       | Jack      | 0     | 18         | css,css      | technology,technology  |
| 4       | James     | 1     | 5          | html         | political              |
| 2       | Peter     | 0     | 0          | null         | null                   |
| 3       | Ali       | 0     | 0          | null         | null                   |
+---------+-----------+-------+------------+--------------+------------------------+

看到？css,css,technology, technology.为什么这些是重复的？我只是增加了一个LEFT JOIN对categories,一模一样tags.但它不能按预期工作,甚至也会对标签产生影响.

---

无论如何,这是预期的结果:

+---------+-----------+-------+------------+--------------+------------------------+
| user_id | user_name | score | reputation | top_two_tags |        category        |
+---------+-----------+-------+------------+--------------+------------------------+
| 1       | Jack      | 0     | 18         | css,mysql    | technology,social      |
| 4       | James     | 1     | 5          | html         | political              |
| 2       | Peter     | 0     | 0          | null         | null                   |
| 3       | Ali       | 0     | 0          | null         | null                   |
+---------+-----------+-------+------------+--------------+------------------------+

有谁知道我怎么能做到这一点？

---

CREATE TABLE users(id integer PRIMARY KEY, user_name varchar(5));
CREATE TABLE tags(id integer NOT NULL PRIMARY KEY, tag varchar(5));
CREATE TABLE reputations(
    id  integer PRIMARY KEY, 
    post_id  integer /* REFERENCES posts(id) */, 
    user_id integer REFERENCES users(id), 
    score integer, 
    reputation integer, 
    date_time integer);
CREATE TABLE post_tag(
    post_id integer /* REFERENCES posts(id) */, 
    tag_id integer REFERENCES tags(id),
    PRIMARY KEY (post_id, tag_id));
CREATE TABLE categories(id INTEGER NOT NULL PRIMARY KEY, category varchar(10) NOT NULL);
CREATE TABLE post_category(
    post_id INTEGER NOT NULL /* REFERENCES posts(id) */, 
    category_id INTEGER NOT NULL REFERENCES categories(id),
    PRIMARY KEY(post_id, category_id)) ;

SELECT
    q1.user_id, q1.user_name, q1.score, q1.reputation, 
    substring_index(group_concat(q2.tag  ORDER BY q2.tag_reputation DESC SEPARATOR ','), ',', 2) AS top_two_tags,
    substring_index(group_concat(q3.category  ORDER BY q3.category_reputation DESC SEPARATOR ','), ',', 2) AS category
FROM
    (SELECT 
        u.id AS user_Id, 
        u.user_name,
        coalesce(sum(r.score), 0) as score,
        coalesce(sum(r.reputation), 0) as reputation
    FROM 
        users u
        LEFT JOIN reputations r 
            ON    r.user_id = u.id 
              AND r.date_time > 1500584821 /* unix_timestamp(DATE_SUB(now(), INTERVAL 1 WEEK)) */
    GROUP BY 
        u.id, u.user_name
    ) AS q1
    LEFT JOIN
    (
    SELECT
        r.user_id AS user_id, t.tag, sum(r.reputation) AS tag_reputation
    FROM
        reputations r 
        JOIN post_tag pt ON pt.post_id = r.post_id
        JOIN tags t ON t.id = pt.tag_id
    WHERE
        r.date_time > 1500584821 /* unix_timestamp(DATE_SUB(now(), INTERVAL 1 WEEK)) */
    GROUP BY
        user_id, t.tag
    ) AS q2
    ON q2.user_id = q1.user_id 
    LEFT JOIN
    (
    SELECT
        r.user_id AS user_id, c.category, sum(r.reputation) AS category_reputation
    FROM
        reputations r 
        JOIN post_category ct ON ct.post_id = r.post_id
        JOIN categories c ON c.id = ct.category_id
    WHERE
        r.date_time > 1500584821 /* unix_timestamp(DATE_SUB(now(), INTERVAL 1 WEEK)) */
    GROUP BY
        user_id, c.category
    ) AS q3
    ON q3.user_id = q1.user_id 
GROUP BY
    q1.user_id, q1.user_name, q1.score, q1.reputation
ORDER BY
    q1.reputation DESC, q1.score DESC ;

Answer 1

您的第二个查询的形式为：

q1 -- PK user_id
LEFT JOIN (...
    GROUP BY user_id, t.tag
) AS q2
ON q2.user_id = q1.user_id 
LEFT JOIN (...
    GROUP BY user_id, c.category
) AS q3
ON q3.user_id = q1.user_id
GROUP BY -- group_concats

内部 GROUP BY 导致(user_id, t.tag)&(user_id, c.category)成为键/UNIQUE。除此之外我不会讨论那些 GROUP BY。

TL;DR当您将 (q1 JOIN q2) 连接到 q3 时，它不在其中一个键/唯一键上，因此对于每个 user_id，您都会为标签和类别的每种可能组合获得一行。因此，最终的 GROUP BY 输入每个（user_id，标签）和每个（user_id，类别）的重复项，并且不恰当地 GROUP_CONCAT 每个 user_id 的重复标签和类别。正确的是 (q1 JOIN q2 GROUP BY) JOIN (q1 JOIN q3 GROUP BY)，其中所有联接都在公共键/UNIQUE 上(user_id)，并且没有虚假聚合。尽管有时您可以撤消这种虚假聚合。

正确的对称 INNER JOIN 方法： LEFT JOIN q1 & q2--1:many--then GROUP BY & GROUP_CONCAT （这就是您的第一个查询所做的）；然后分别类似地 LEFT JOIN q1 & q3--1:many--然后 GROUP BY & GROUP_CONCAT; 然后 INNER JOIN 将两个结果 ON user_id--1:1。

正确的对称标量子查询方法：从 q1 中选择 GROUP_CONCAT 作为标量子查询，每个子查询都带有 GROUP BY。

正确的累积LEFT JOIN方法：LEFT JOIN q1 & q2--1:many--then GROUP BY & GROUP_CONCAT; 然后 LEFT JOIN that & q3--1:many--然后 GROUP BY & GROUP_CONCAT。

像您的第二个查询一样的正确方法：您首先 LEFT JOIN q1 & q2--1:many。然后你离开加入 & q3--many:1:many。它为与 user_id 一起出现的标签和类别的每种可能组合提供一行。然后，在 GROUP BY 后，通过 GROUP_CONCAT - 重复的（user_id，标签）对和重复的（user_id，类别）对。这就是为什么你有重复的列表元素。但是将 DISTINCT 添加到 GROUP_CONCAT 会给出正确的结果。（佩尔奇基托的评论。）

像往常一样，您更喜欢的是根据实际数据/使用情况/统计信息通过查询计划和计时来了解的工程权衡。一个问题是，many:1:many JOIN 方法的额外行是否抵消了 GROUP BY 的节省。

-- cumulative LEFT JOIN approach
SELECT
   q1.user_id, q1.user_name, q1.score, q1.reputation,
    top_two_tags,
    substring_index(group_concat(q3.category  ORDER BY q3.category_reputation DESC SEPARATOR ','), ',', 2) AS category
FROM
    -- your 1st query (less ORDER BY) AS q1
    (SELECT
        q1.user_id, q1.user_name, q1.score, q1.reputation, 
        substring_index(group_concat(q2.tag  ORDER BY q2.tag_reputation DESC SEPARATOR ','), ',', 2) AS top_two_tags
    FROM
        (SELECT 
            u.id AS user_Id, 
            u.user_name,
            coalesce(sum(r.score), 0) as score,
            coalesce(sum(r.reputation), 0) as reputation
        FROM 
            users u
            LEFT JOIN reputations r 
                ON    r.user_id = u.id 
                  AND r.date_time > 1500584821 /* unix_timestamp(DATE_SUB(now(), INTERVAL 1 WEEK)) */
        GROUP BY 
            u.id, u.user_name
        ) AS q1
        LEFT JOIN
        (
        SELECT
            r.user_id AS user_id, t.tag, sum(r.reputation) AS tag_reputation
        FROM
            reputations r 
            JOIN post_tag pt ON pt.post_id = r.post_id
            JOIN tags t ON t.id = pt.tag_id
        WHERE
            r.date_time > 1500584821 /* unix_timestamp(DATE_SUB(now(), INTERVAL 1 WEEK)) */
        GROUP BY
            user_id, t.tag
        ) AS q2
        ON q2.user_id = q1.user_id 
        GROUP BY
            q1.user_id, q1.user_name, q1.score, q1.reputation
    ) AS q1
    -- finish like your 2nd query
    LEFT JOIN
    (
    SELECT
        r.user_id AS user_id, c.category, sum(r.reputation) AS category_reputation
    FROM
        reputations r 
        JOIN post_category ct ON ct.post_id = r.post_id
        JOIN categories c ON c.id = ct.category_id
    WHERE
        r.date_time > 1500584821 /* unix_timestamp(DATE_SUB(now(), INTERVAL 1 WEEK)) */
    GROUP BY
        user_id, c.category
    ) AS q3
    ON q3.user_id = q1.user_id 
GROUP BY
    q1.user_id, q1.user_name, q1.score, q1.reputation
ORDER BY
    q1.reputation DESC, q1.score DESC ;