合并到一个具有相同 id 或属性 SQL Group_concat 的组没有帮助:(

Posted 2023-02-24

【中文标题】合并到一个具有相同 id 或属性 SQL Group_concat 的组没有帮助:(

【英文标题】：Combine to one group with the same id or attribute SQL Group_concat doesn't help:(

【发布时间】：2021-06-18 08:37:57

【问题描述】：

如果它们具有相同的 id 或相同的属性，我需要将它们组合在一行中。所以我想我需要使用INNER JOIN 和GROUP_CONCAT，但我不知道如何使用。问题是，如果两个用户没有共同的属性，但与相同的第三个用户属于同一个组，那么这三个用户必须合并为一个组。另外我的表中没有group_id 列。

group_id, user_id, group_attributes
1, 1, "red, green, yellow, grey, purple, coffeemaker"
1, 2, "red, green, yellow, grey, purple, coffeemaker"
1, 3, "red, green, yellow, grey, purple, coffeemaker"
1, 4, "red, green, yellow, grey, purple, coffeemaker"
1, 5, "red, green, yellow, grey, purple, coffeemaker"
2, 6, "coffee, milk, croissant"
2, 7, "coffee, milk, croissant"
2, 8, "coffee, milk, croissant"

原始数据可缩短您的回答时间。

CREATE TABLE task (
    user_id INT(10) NOT NULL,
    attribute VARCHAR(50) NULL DEFAULT NULL);
INSERT INTO task (user_id, attribute)
VALUES
(1, 'red'),
(1, 'green'),
(2, 'green'),
(2, 'yellow'),
(3, 'grey'),
(3, 'coffeemaker'),
(4, 'grey'),
(4, 'purple'),
(5, 'purple'),
(5, 'red'),
(6, 'black'),
(7, 'black'),
(7, 'milk'),
(8, 'milk'),
(8, 'croissant');

【参考方案1】：

这是一个图行走问题，所以一个简单的JOIN 是不够的。一种方法是获取与给定属性关联的所有属性。以下递归 CTE 执行此操作：

with recursive aa as (
      select distinct t1.attribute as at1, t2.attribute as at2
      from task t1 join
           task t2
           on t1.user_id = t2.user_id
     ),
     cte as (
      select at1, at2, at1 as found, 1 as lev
      from aa
      union all
      select cte.at1, aa.at2, concat_ws(',', found, aa.at2), lev + 1
      from cte join
           aa
           on cte.at2 = aa.at1
      where find_in_set(aa.at2, found) = 0 
     )
select distinct at1, at2
from cte;

然后您可以使用相同的递归 CTE 将值组合成一个字符串：

with recursive aa as (
      select distinct t1.attribute as at1, t2.attribute as at2
      from task t1 join
           task t2
           on t1.user_id = t2.user_id
     ),
     cte as (
      select at1, at2, at1 as found, 1 as lev
      from aa
      union all
      select cte.at1, aa.at2, concat_ws(',', found, aa.at2), lev + 1
      from cte join
           aa
           on cte.at2 = aa.at1
      where find_in_set(aa.at2, found) = 0 
     )
select dense_rank() over (order by pairs.all_attributes) as group_id, t.user_id, pairs.all_attributes
from (select at1, group_concat(at2) as all_attributes
      from cte
      group by at1
     ) pairs join
     (select user_id, min(attribute) as min_attribute
      from task
      group by user_id
     ) t
     on t.min_attribute = pairs.at1;

我看不出这段代码有什么问题。但是 dbfiddle 坚持为pairs 创建一个十六进制字符串。但是，我认为这将适用于您的数据库。这是fiddle。