SQL 根据另一个表查找缺少列的行

问题描述

我有一个场景，我需要使用一个有效元素表来查找另一个表中缺少这些元素的记录，以便我可以反过来修补它们。这有点棘手，因为缺少的列数据在“嵌套”列（例如，它在另一列中具有重复条目）。我觉得某种 LEFT OUTER JOIN 可能在这里起作用，但我不能完全理解。 FWIW，我的基准是 Oracle 19c，但我最终也需要支持 Postgres 11+。问题总结为：

给定租户表：

租户
1000
2000
3000

还有一个组表，其中每一天对应一个新的组id，每个组有一对多的租户，其中每个租户对应一个子组id：

日期	组	租户	子组
2021-02-16	G1	1000	SG1
2021-02-16	G1	2000	SG2
2021-02-16	G1	3000	SG3
2021-02-17	G2	1000	SG4
2021-02-17	G2	2000	SG5
2021-02-18	G3	2000	SG6
2021-02-18	G3	3000	SG7
2021-02-19	G4	1000	SG8

查找缺少租户的组：

组	租户
G2	3000
G3	1000
G4	2000
G4	3000

解决方法

WITH
   tab1 AS
      ( SELECT TO_DATE('16.02.2021','DD.MM.YYYY') AS date_col,'G1' AS group_col,1000 AS tenant,'SG1' AS subgroup FROM DUAL
        UNION ALL
        SELECT TO_DATE('16.02.2021',2000 AS tenant,'SG2' AS subgroup FROM DUAL
        UNION ALL
        SELECT TO_DATE('16.02.2021',3000 AS tenant,'SG3' AS subgroup FROM DUAL
        UNION ALL
        SELECT TO_DATE('17.02.2021','G2' AS group_col,'SG4' AS subgroup FROM DUAL
        UNION ALL
        SELECT TO_DATE('17.02.2021','SG5' AS subgroup FROM DUAL
        UNION ALL
        SELECT TO_DATE('18.02.2021','G3' AS group_col,'SG6' AS subgroup FROM DUAL
        UNION ALL
        SELECT TO_DATE('18.02.2021','SG7' AS subgroup FROM DUAL
        UNION ALL
        SELECT TO_DATE('19.02.2021','G4' AS group_col,'SG8' AS subgroup FROM DUAL
      ),tab2 AS
      ( SELECT 1000 AS tenant FROM DUAL
        UNION ALL
        SELECT 2000 AS tenant FROM DUAL
        UNION ALL
        SELECT 3000 AS tenant FROM DUAL
      )
SELECT *
  FROM ( SELECT t1.group_col,t2.tenant
           FROM ( SELECT DISTINCT group_col FROM tab1) t1
           CROSS JOIN tab2 t2
       ) x
 WHERE NOT EXISTS ( SELECT *
                      FROM tab1
                     WHERE tab1.group_col = x.group_col
                       AND tab1.tenant = x.tenant
                  )
  ORDER BY group_col,tenant;

结果：

GR     TENANT
-- ----------
G2       3000
G3       1000
G4       2000
G4       3000

该查询也适用于 Postgres（然后删除 FROM DUAL，否则您将使其适应您的表格）。

,

在 Oracle 中，您可以使用 partitioned outer join 来填充缺失的行，然后您可以仅选择这些行

select g.group_col,t.tenant
  from groups g partition by (group_col)
 right outer join tenants t on t.tenant = g.tenant
 where g.tenant is null

,

我通过使用 cross join 生成所有行然后删除存在的行来解决这个问题。以下是标准 SQL，适用于 Oracle 和 Postgres：

select g.group_id,t.tenant
from tenants t cross join
     (select distinct group_id from groups) g left join
     groups g2
     using (tenant,group_id)
where g2.tenant is null;