如何比较多行

Question

我有一个 table 的数据如下，想要 return 那些 group_id 具有独特的数据。 group_id 3 和 4 都有两个分量 123 和 456，所以它们是“重复的”，我们只需要 return 较小的 group_id，即 3。还有 group_id 5没有重复，可以 returned。所以我们希望 group_id 3 和 5 被 returned.

如何针对 postgres 数据库编写 SQL 查询来实现该目的？谢谢！

id	group_id	component_id
1	3	123
2	3	456
3	4	123
4	4	456
5	5	123

Answer 1

SELECT group_id, MIN(component_id)
FROM   MyTable
GROUP  BY group_id
HAVING COUNT(*) > 1

Answer 2

这是一种将 group_id 分配给 component_id 的方法。

它使用带数组的递归 CTE 来查找可能的组合。
递归从孤独的group_id开始。

然后下一个 CTE 选择最长的组合之一。

WITH RECURSIVE RCTE AS (
    SELECT id, group_id, component_id
    , 1 as Lvl
    , array[group_id] as group_ids
    , array[component_id] as component_ids
    FROM YourTable
    WHERE group_id IN (
      SELECT group_id
      FROM YourTable
      GROUP BY group_id
      HAVING COUNT(*) = 1
    )
    UNION ALL
    SELECT t.id, t.group_id, t.component_id
    , Lvl+1
    , cte.group_ids || t.group_id
    , cte.component_ids || t.component_id
    FROM RCTE cte
    JOIN YourTable t 
      ON t.group_id != ALL(group_ids)
     AND t.component_id != ALL(component_ids)
)
, CTE_ARRAYS AS (
    SELECT group_ids, component_ids
    FROM RCTE
    ORDER BY array_length(group_ids, 1) desc, Lvl desc
    LIMIT 1
) 
SELECT a.group_id, a.component_id
FROM CTE_ARRAYS c
CROSS JOIN LATERAL UNNEST(c.group_ids, c.component_ids) WITH ORDINALITY AS a(group_id, component_id)
ORDER BY a.group_id;

group_id	component_id
3	456
5	123

db<>fiddle here

Answer 3

使用 2 级聚合：

SELECT MIN(group_id) group_id
FROM (
  SELECT group_id, STRING_AGG(component_id::text, ',' ORDER BY component_id) components
  FROM tablename
  GROUP BY group_id
) t
GROUP BY components;

参见demo。

如何比较多行

How to compare multiple rows

sql

postgresql

group-by

min

string-agg