SQL中Cte中的Cte

我遇到了一种情况，需要对CTE中的CTE结果应用where、group by条件。

表1如下所示

+---+---+---+---+
| x | y | z | w |
+---+---+---+---+
| 1 | 2 | 3 | 1 |
| 2 | 3 | 4 | 2 |
| 3 | 2 | 5 | 3 |
| 1 | 2 | 6 | 2 |
+---+---+---+---+

+---+---+-----+---+
| a | b |  c  | d |
+---+---+-----+---+
| 1 | m | 100 | 1 |
| 2 | n |  23 | 2 |
| 4 | o |  34 | 4 |
| 1 | m |  23 | 2 |
+---+---+-----+---+

+---+---+---+---+---+-----+
| x | y | z | w | b |  c  |
+---+---+---+---+---+-----+
| 1 | 2 | 3 | 1 | m | 100 |
| 2 | 3 | 4 | 2 | n |  23 |
| 1 | 2 | 6 | 2 | m |  23 |
+---+---+---+---+---+-----+

+---+---+---+---+---+-----+
| x | y | z | w | b |  c  |
+---+---+---+---+---+-----+
| 1 | 2 | 3 | 1 | m | 100 |
| 1 | 2 | 6 | 2 | m |  23 |
+---+---+---+---+---+-----+

表2如下所示

+---+---+---+---+
| x | y | z | w |
+---+---+---+---+
| 1 | 2 | 3 | 1 |
| 2 | 3 | 4 | 2 |
| 3 | 2 | 5 | 3 |
| 1 | 2 | 6 | 2 |
+---+---+---+---+

+---+---+-----+---+
| a | b |  c  | d |
+---+---+-----+---+
| 1 | m | 100 | 1 |
| 2 | n |  23 | 2 |
| 4 | o |  34 | 4 |
| 1 | m |  23 | 2 |
+---+---+-----+---+

+---+---+---+---+---+-----+
| x | y | z | w | b |  c  |
+---+---+---+---+---+-----+
| 1 | 2 | 3 | 1 | m | 100 |
| 2 | 3 | 4 | 2 | n |  23 |
| 1 | 2 | 6 | 2 | m |  23 |
+---+---+---+---+---+-----+

+---+---+---+---+---+-----+
| x | y | z | w | b |  c  |
+---+---+---+---+---+-----+
| 1 | 2 | 3 | 1 | m | 100 |
| 1 | 2 | 6 | 2 | m |  23 |
+---+---+---+---+---+-----+

假设我在一个名为TAB的表中有以下sql查询的数据

with cte as (
select x,y,z from table1),
cte1 as (select a,b,c from table2)
select cte.x,cte1.y,cte1.z,cte2.b,cte2.c from cte left join cte1 on cte.x=cte.a and cte1.w=cte2.d

上述CTE的结果如下

+---+---+---+---+
| x | y | z | w |
+---+---+---+---+
| 1 | 2 | 3 | 1 |
| 2 | 3 | 4 | 2 |
| 3 | 2 | 5 | 3 |
| 1 | 2 | 6 | 2 |
+---+---+---+---+

+---+---+-----+---+
| a | b |  c  | d |
+---+---+-----+---+
| 1 | m | 100 | 1 |
| 2 | n |  23 | 2 |
| 4 | o |  34 | 4 |
| 1 | m |  23 | 2 |
+---+---+-----+---+

+---+---+---+---+---+-----+
| x | y | z | w | b |  c  |
+---+---+---+---+---+-----+
| 1 | 2 | 3 | 1 | m | 100 |
| 2 | 3 | 4 | 2 | n |  23 |
| 1 | 2 | 6 | 2 | m |  23 |
+---+---+---+---+---+-----+

+---+---+---+---+---+-----+
| x | y | z | w | b |  c  |
+---+---+---+---+---+-----+
| 1 | 2 | 3 | 1 | m | 100 |
| 1 | 2 | 6 | 2 | m |  23 |
+---+---+---+---+---+-----+

我想从表选项卡查询以下内容

select * from TAB where (X||b) in (select (X||b) from TAB group by (X||Y) having sum(c)=123)

我正试图按如下方式表述SQL查询，但它并不像我预期的那样：

select * from (

with cte as (
select x,y,z from table1),
cte1 as (select a,b,c from table2)
select cte.x,cte1.y,cte1.z,cte2.b,cte2.c from cte left join cte1 on cte.x=cte.a) as TAB 

where ((X||b) in (select (X||b) from TAB group by (X||Y) having sum(c)=123))

最终结果必须如下所示

+---+---+---+---+
| x | y | z | w |
+---+---+---+---+
| 1 | 2 | 3 | 1 |
| 2 | 3 | 4 | 2 |
| 3 | 2 | 5 | 3 |
| 1 | 2 | 6 | 2 |
+---+---+---+---+

+---+---+-----+---+
| a | b |  c  | d |
+---+---+-----+---+
| 1 | m | 100 | 1 |
| 2 | n |  23 | 2 |
| 4 | o |  34 | 4 |
| 1 | m |  23 | 2 |
+---+---+-----+---+

+---+---+---+---+---+-----+
| x | y | z | w | b |  c  |
+---+---+---+---+---+-----+
| 1 | 2 | 3 | 1 | m | 100 |
| 2 | 3 | 4 | 2 | n |  23 |
| 1 | 2 | 6 | 2 | m |  23 |
+---+---+---+---+---+-----+

+---+---+---+---+---+-----+
| x | y | z | w | b |  c  |
+---+---+---+---+---+-----+
| 1 | 2 | 3 | 1 | m | 100 |
| 1 | 2 | 6 | 2 | m |  23 |
+---+---+---+---+---+-----+

---

我认为DB2不允许在子查询中使用CTE，也不允许嵌套CTE。为什么不直接用另一个CTE来写呢

with cte as (
      select x,y,z from
      table1
     ),
     cte1 as (
      select a,b,c
      from table2
     ),
     tab as (
      select cte.x,cte1.y,cte1.z,cte1.w,cte2.b,cte2.c
      from cte left join
           cte1
           on cte.x=cte.a and cte1.w=cte2.d
     )
select *
from TAB
where (X||b) in (select (X||b) from TAB group by (X||Y) having sum(c)=123);

---

输入数据是什么，预期结果是什么？通常，你不需要那种复杂性。你可以嵌套表达式。CTE的用途是什么？@data\u henrik实际上，table1和table2查询包含交叉连接和联合所有选项，因此需要CTE来连接所有这些结果。据我所知，你根本不需要CTE，你只是连接了两个集合