SQL 根据另一个表查找缺少列的行答案

【问题标题】：SQL Find rows with missing columns based on another tableSQL 根据另一个表查找缺少列的行
【发布时间】：2021-05-22 01:58:36
【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，但我最终还需要支持 Postgres 11+。问题总结为：

给定一个租户表：

Tenant
1000
2000
3000

还有一个groups表，其中每一天对应一个新的group id，每个group有一对多租户，其中每个tenent对应一个subgroup id：

Date	Group	Tenant	Subgroup
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：

Group	Tenant
G2	3000
G3	1000
G4	2000
G4	3000

【问题讨论】：

标签： sql postgresql oracle postgresql-11 oracle19c

【解决方案1】：

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', 'DD.MM.YYYY') AS date_col, 'G1' AS group_col, 2000 AS tenant, 'SG2' AS subgroup FROM DUAL
        UNION ALL
        SELECT TO_DATE('16.02.2021', 'DD.MM.YYYY') AS date_col, 'G1' AS group_col, 3000 AS tenant, 'SG3' AS subgroup FROM DUAL
        UNION ALL
        SELECT TO_DATE('17.02.2021', 'DD.MM.YYYY') AS date_col, 'G2' AS group_col, 1000 AS tenant, 'SG4' AS subgroup FROM DUAL
        UNION ALL
        SELECT TO_DATE('17.02.2021', 'DD.MM.YYYY') AS date_col, 'G2' AS group_col, 2000 AS tenant, 'SG5' AS subgroup FROM DUAL
        UNION ALL
        SELECT TO_DATE('18.02.2021', 'DD.MM.YYYY') AS date_col, 'G3' AS group_col, 2000 AS tenant, 'SG6' AS subgroup FROM DUAL
        UNION ALL
        SELECT TO_DATE('18.02.2021', 'DD.MM.YYYY') AS date_col, 'G3' AS group_col, 3000 AS tenant, 'SG7' AS subgroup FROM DUAL
        UNION ALL
        SELECT TO_DATE('19.02.2021', 'DD.MM.YYYY') AS date_col, 'G4' AS group_col, 1000 AS tenant, '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

查询也可以在 Postgres 中使用（然后删除 FROM DUAL 否则您会将其调整到您的表中）。

【讨论】：

【解决方案2】：

在 Oracle 中，您可以使用 partitioned outer join 来填写缺失的行，然后您可以只选择这些行

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

【讨论】：

【解决方案3】：

我通过使用cross join 来生成所有行，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;

【讨论】：