CTE 不检索任何行用于左连接和 where 表达式

Question

我在 Amazon Redshift 上写了一个 SQL 查询，我遇到了一个 CTE 问题，return 没有行。

在比较 CTE No1 的时间戳字段与 CTE No2 的时间戳字段之后，查询正在从 CTE No1 和 returns 字段中检索 ID 和时间戳，并从 CTE No 2,3,4,5,6 中检索字段， 3、4、5。更准确地说，下面是我的 SQL 查询的伪代码，我希望它是可以理解的。请不要过多关注 LEAD() 和 LAG() 函数查询问题本质上比它可能出现的要简单得多，并且是关于 CTE 何时有 0 行，但它们用于最终的 where 条件查询，如何解释这些情况。

with CTE1 AS 
(SELECT timestamp1, id
from table1
where ...
...
...)

, CTE2 as 
(select timestamp2, 
        id,  
        field1, 
        field2, 
        LAG(timestamp2,1) over partition by id order by timestamp2 as timestamp2_previous, 
        LEAD(timestamp2,1) over partition by id order by timestamp2 as timestamp2_next 
        from table2)

, CTE3 as 
(select timestamp3, 
        id, 
        field1, 
        field2, 
        LAG(timestamp3,1) over partition by id order by timestamp3 as timestamp3_previous, 
        LEAD(timestamp3,1) over partition by id order by timestamp3 as timestamp3_next 
       from table3)

, CTE4 as 
(select timestamp4, 
        id, 
        field1, 
        field2, 
        LAG(timestamp4,1) over partition by id order by timestamp4 as timestamp4_previous, 
        LEAD(timestamp4,1) over partition by id order by timestamp4 as timestamp4_next 
        from table4)

, CTE5 as 
(select timestamp5, 
        id, 
        field1, 
        field2, 
        LAG(timestamp5,1) over partition by id order by timestamp5 as timestamp5_previous, 
        LEAD(timestamp5,1) over partition by id order by timestamp5 as timestamp5_next 
        from table5)

, CTE6 as 
(select timestamp6, 
        id, 
        field1, 
        field2, 
        LAG(timestamp6,1) over partition by id order by timestamp6 as timestamp6_previous, 
        LEAD(timestamp6,1) over partition by id order by timestamp6 as timestamp6_next 
        from table6)

select cte1.id, 
       cte1.timestamp1, 
       case when cte1.timestamp1<cte2.timestamp2 and cte2.timestamp2_previous is null then cte2.field1 else cte2.field2, 
case when cte1.timestamp1<cte3.timestamp3 and cte3.timestamp3_previous is null then cte3.field1 else cte3.field2, 
case when cte1.timestamp1<cte4.timestamp4 and cte4.timestamp4_previous is null then cte4.field1 else cte4.field2, 
case when cte1.timestamp1<cte5.timestamp5 and cte5.timestamp5_previous is null then cte5.field1 else cte5.field2, 
case when cte1.timestamp1<cte6.timestamp6 and cte6.timestamp6_previous is null then cte6.field1 else cte6.field2

from cte1
left join cte2 on cte1.id=cte2.id
left join cte3 on cte1.id=cte3.id
left join cte4 on cte1.id=cte4.id
left join cte5 on cte1.id=cte5.id
left join cte6 on cte1.id=cte6.id
where 1 

and 

-- conditions for CTE2

( case when cte1.timestamp1 < cte2.timestamp2 and cte2.timestamp2_previous is null then 1=1
         else cte1.timestamp1 > cte2.timestamp2 end
and
        CASE
                WHEN cte2.timestamp2_next is not null THEN cte1.timestamp1 < cte2.timestamp2_next ELSE 1=1
        END
)

-- conditions for CTE3

( case when cte1.timestamp1 < cte3.timestamp3 and cte3.timestamp3_previous is null then 1=1
         else cte1.timestamp1 > cte3.timestamp3 end
and
        CASE
                WHEN cte3.timestamp3_next is not null THEN cte1.timestamp1 < cte3.timestamp3_next ELSE 1=1
        END
)

-- conditions for CTE4

( case when cte1.timestamp1 < cte4.timestamp4 and cte4.timestamp4_previous is null then 1=1
         else cte1.timestamp1 > cte4.timestamp4 end
and
        CASE
                WHEN cte4.timestamp4_next is not null THEN cte1.timestamp1 < cte4.timestamp4_next ELSE 1=1
        END
)


-- conditions for CTE5

( case when cte1.timestamp1 < cte5.timestamp5 and cte5.timestamp5_previous is null then 1=1
         else cte1.timestamp1 > cte5.timestamp5 end
and
        CASE
                WHEN cte5.timestamp5_next is not null THEN cte1.timestamp1 < cte5.timestamp5_next ELSE 1=1
        END
)


-- conditions for CTE6

( case when cte1.timestamp1 < cte6.timestamp6 and cte6.timestamp6_previous is null then 1=1
         else cte1.timestamp1 > cte6.timestamp6 end
and
        CASE
                WHEN cte6.timestamp6_next is not null THEN cte1.timestamp1 < cte6.timestamp6_next ELSE 1=1
        END
)

问题是第一个之后的一些 CTE，return 没有特定 ID 的记录（我针对特定 ID 测试了我的代码，而 CTE3 return 没有该 ID 的行). 在这些情况下，最后的 where 条件，它们都与 AND 连接，将评估其中没有记录的 CTE，并且 return 0 行（因为操作 TRUE AND NULL），即使所有其他 CTE 有该 ID 的数据，查询结果应该是为它们检索数据。

如何检索此类情况的结果，并在引用具有当前行的 CTE 的结果字段中获取 NULL？

Answer 1

正如@JNevill 所说，答案是将 where 条件移动到连接中。为了让您了解这是如何工作的，您在 cte1 和 cte2 之间的连接变为： ON cte1.id = cte2.id AND ((cte1.timestamp < cte2.timestamp and cte2.timestamp_previous IS NULL) OR cte1.timestamp > cte2.timestamp) AND (cte2.timestamp_next IS NULL OR cte1.timestamp < cte2.timestamp_next)

CTE 不检索任何行用于左连接和 where 表达式

CTE retrieving no rows is used in left join and where expression

sql

null

common-table-expression

amazon-redshift