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

作者:互联网

Here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here是我的新查询.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,技术,技术.为什么这些是重复的?我刚刚为类别添加了一个LEFT JOIN,就像标签一样.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 ;

解决方法:

您的第二个查询是以下形式:

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)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组合组合获得一行.类别.所以最终的GROUP BY输入重复(user_id,tag)& per(user_id,category)和不恰当的GROUP_CONCATs重复标签&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)&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& q2–1:很多 – 然后GROUP BY& GROUP_CONCAT(这是你的第一个查询所做的);然后分别类似LEFT JOIN q1& q3–1:很多 – 然后GROUP BY& 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将q1中的GROUP_CONCATs选为scalar subqueries.

正确的累积LEFT JOIN方法:LEFT JOIN q1& q2–1:很多 – 然后GROUP BY& GROUP_CONCAT;然后LEFT JOIN& q3–1:很多 – 然后GROUP BY& GROUP_CONCAT.

像你的第二个查询一样正确的方法:你首先LEFT JOIN q1& q2–1:很多.然后你LEFT JOIN& 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. (根据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&时间,每个实际数据/使用/统计.输入和输入预期复制量的统计数据),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 ;

标签:sql,mysql,group-by,group-concat,left-join
来源: https://codeday.me/bug/20190916/1807048.html