首页 > 解决方案 > 计算 MYSQL 中每个用户的子表中的不同记录

问题描述

我有一个比赛,计算每个用户收集了多少物种。这由 3 个表管理:

  1. 一个名为“sub”的父表带有集合,每个集合都是唯一的,有一个 id 并与一个用户 id 相关联。
    
    +----+---------+
    | id | user_id |
    +----+---------+
    |  1 |       1 |
    |  2 |      10 |
    |  3 |       1 |
    |  4 |       3 |
    |  5 |       1 |
    |  6 |      10 |
    +----+---------+
    
  2. 名为“sub_items”的子表包含规范的多个唯一记录,并通过子ID到ID与父表相关。(每个子可以有多个规范记录)

+----+--------+---------+--+
| id | sub_id | spec_id |  |
+----+--------+---------+--+
|  1 |      1 |    1000 |  |
|  2 |      1 |    1003 |  |
|  3 |      1 |    2520 |  |
|  4 |      2 |    7600 |  |
|  5 |      2 |    1000 |  |
|  6 |      3 |      15 |  |
+----+--------+---------+--+

  1. 具有关联 user_id 的用户表

+--------+-------+--+
| usename | name    |
+---------+-------+--+
|      1 | David    |
|     10 | Ruth     |
|      3 | Rick     |
+--------+-------+--+

我需要按降序列出具有最独特规格的用户。预期输出:David 共有 2 个独特的规格。Ruth 共有 2 个独特的规格。

+--------+---------+
| id     | total   |
+----+-------------+
|  David |    2    |
|  Ruth  |    2    |
|  Rick  |    2    |
+----+-------------+

到目前为止,我有这个,它产生了一个结果。但它不准确,它计算总记录。我可能在子查询中的某处遗漏了一个 DISTINCT。

SELECT s.id, s.user_id,u.name, sum(t.count) as total
FROM sub s
LEFT JOIN (
    SELECT id, sub_id, count(id) as count FROM sub_items GROUP BY sub_id
 ) t ON t.sub_id = s.id
 LEFT JOIN user u ON u.username = s.user_id
 GROUP BY user_id
 ORDER BY total DESC
 

我看过这个解决方案,但它没有考虑独特的方面

标签: mysqldistinct

解决方案


您首先必须获得所有用户的最大“分数”,例如:

    SELECT count(DISTINCT si.id) as total
    FROM sub INNER JOIN sub_items si ON sub.id = su.sub_id
    GROUP BY sub.user_id
    ORDER BY total DESC
    LIMIT 1

然后,您可以使用它将查询限制为共享该最高分数的用户:

SELECT u.name, count(DISTINCT si.id) as total
FROM
    user u
    INNER JOIN sub ON u.usename = sub.user_id
    INNER JOIN sub_items si ON sub.id = su.sub_id
GROUP BY u.name
HAVING total =
    (
        SELECT count(DISTINCT si.id) as total
        FROM sub INNER JOIN sub_items si ON sub.id = su.sub_id
        GROUP BY sub.user_id
        ORDER BY total DESC
        LIMIT 1
    )

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