嵌套common-table-expression的语法

Question

昨天我问了一个复杂查询的问题，我得到了一个例子。我真的想让它工作，但它有一个语法错误，我无法弄清楚。请记住，本周早些时候我刚刚介绍过CTE，所以希望这很简单。

我认为我不需要在这里发布完整的代码，所以我只是总结一下

 with cte as (select dateadd(hour, 1, cast(cast(getdate() -1 as date) as datetime)) as midnnight),
 allhours as (
 select 0 as hour, midnight as timestart, dateadd(hour, 1, timestart) as timeend from cte union all
 select 1 as hour, dateadd(hour, 1, midnight), dateadd(hour, 2, midnight) from cte union all
 ....
 select 23 as hour, dateadd(hour, 23, midnight), dateadd(hour, 24, midnight) from cte union all
) 
select ah.hour,...

（...）表示我省略的不必要的代码，使其不那么混乱

但是我在select 23和select ah.hour之间的括号上出现语法错误 “语法不正确'）'。期待SELECT，或'（'。

非常感谢任何帮助。

-J

Answer 1

您有一些语法错误，包括底部不需要的UNION ALL，而SELECT中的第一个allhours引用了别名，请尝试以下操作：

;with cte as 
(
  select dateadd(hour, 1, cast(cast(getdate() -1 as date) as datetime)) as midnight
),
allhours as 
(
  select 0 as hour, midnight as timestart, dateadd(hour, 1, midnight) as timeend 
  from cte 
  union all
  select 1 as hour, dateadd(hour, 1, midnight), dateadd(hour, 2, midnight) 
  from cte 
  union all
  select 23 as hour, dateadd(hour, 23, midnight), dateadd(hour, 24, midnight) 
  from cte
) 
select *
from allhours

请参阅SQL Fiddle with Demo

Answer 2

你应该摆脱最后的Union all

 hour, dateadd(hour, 23, midnight), dateadd(hour, 24, midnight) from cte union all 
)  
select ah.hour,... 

另外，拼写在第一行的午夜