Question

Here是我所有表格的结构和查询（请关注最后查询，附后下方）。正如你在小提琴中所看到的，这里是当前输出：

+---------+-----------+-------+------------+--------------+
| 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。 here是我的新查询。正如您在结果中看到的那样，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                   |
+---------+-----------+-------+------------+--------------+------------------------+

请参阅？ css,css，technology, technology。为什么这些是重复的？我刚为LEFT JOIN添加了一个categories，与tags完全相同。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                   |
+---------+-----------+-------+------------+--------------+------------------------+

有谁知道我怎么能实现这个目标？

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

您的第二个查询格式为：

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)＆amp; (user_id, c.category)是键/ UNIQUE。除此之外，我不会解决那些GROUP BY。

TL; DR 当您加入（q1 JOIN q2）到q3时，它不在其中一个的键/ UNIQUE上，因此对于每个user_id，您为每个可能的标签和放大器组合获得一行;类别。所以最终的GROUP BY输入重复（user_id，tag）＆amp; per（user_id，category）和不恰当的GROUP_CONCATs重复标签＆amp;每个user_id的类别。正确的是（q1 JOIN q2 GROUP BY）JOIN（q1 JOIN q3 GROUP BY），其中所有连接都在公共密钥/ UNIQUE (user_id)＆amp;没有虚假的聚合。虽然有时你可以撤消这种虚假聚合。

正确的对称INNER JOIN方法：LEFT JOIN q1＆amp; q2--1：很多 - 然后GROUP BY＆amp; GROUP_CONCAT（这是你的第一个查询所做的）;然后分别类似LEFT JOIN q1＆amp; q3--1：很多 - 然后GROUP BY＆amp; GROUP_CONCAT;然后INNER JOIN两个结果ON user_id - 1：1。

正确的对称标量子查询方法：使用GROUP BY从q1中选择GROUP_CONCAT作为scalar subqueries。

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

像你的第二个查询一样正确的方法：你首先LEFT JOIN q1＆amp; q2--1：很多。然后你LEFT JOIN＆amp; Q3 - 很多：1：很多。它为标签和放大器的每种可能组合提供一排。与user_id一起显示的类别。然后在GROUP BY之后GROUP_CONCAT - 重复（user_id，tag）对和重复（user_id，category）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会得到正确的结果。（按wchiquito的评论。）

您更喜欢的是通常的工程权衡，以便通过查询计划＆amp;时间，每个实际数据/使用/统计。输入和输入预期复制数量的统计数据，实际查询的时间等。一个问题是多个额外的行：1：多JOIN方法是否抵消了它对GROUP BY的保存。

-- 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 ;

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

1 个答案: