Question

我有一个查询，它返回一个结果集，现在我需要修改查询以也返回另一个结果集

我已经运行了下面的代码，并且返回了两个CTE中的tagID的结果，并为您提供了超时时间。我现在需要修改查询，这样它将向我显示所有已扫描的标签，其中包含Time in和out out时间，反之亦然，而没有重复。

With CTE AS
( 
select Tag as 'Tag ID', UID_KEG, CONVERT(VARCHAR(10), MOVEMENT_DATE, 105) 
as [DATE],CONVERT(VARCHAR(10), MOVEMENT_DATE, 108) as [Time Out]
from MOVEMENT M
Inner join KEG K on K.UNIQUE_ID = M.UID_KEG
where Convert(varchar(10),MOVEMENT_DATE,120) = '2019-06-13' 
and UID_STATION = 4
and TAG <> 'NO TAG'
) ,
CTE2 AS
(select Tag as 'Tag ID', UID_KEG, CONVERT(VARCHAR(10), MOVEMENT_DATE, 105) 
as [DATE],CONVERT(VARCHAR(10), MOVEMENT_DATE, 108) as [Time IN]
from MOVEMENT M
Inner join KEG K on K.UNIQUE_ID = M.UID_KEG
where Convert(varchar(10),MOVEMENT_DATE,120) = '2019-06-13'
and UID_STATION = 5
and TAG <> 'NO TAG'
)
Select CTE.[Tag ID], CTE.[DATE], [Time IN], [Time Out],DATEDIFF(MINUTE, 
[Time IN], [Time Out]) as [Time in Process]
from CTE
Inner Join CTE2 on CTE2.[Tag ID] = CTE.[Tag ID]
where Exists (Select CTE2.[Tag ID] 
       from CTE2
       where CTE2.[Tag ID] = CTE.[Tag ID] )

此刻的查询为我提供了以下结果：

TAG ID                      DATE        Time_In     Time_Out    DIF
33154A36D00F46C000007144    6/13/2019   4:43:05 AM  6:25:27 AM  102
33154A36D00F46C00000464A    6/13/2019   4:43:47 AM  6:06:45 AM  83
33154A36D00F46C000006DFF    6/13/2019   4:46:22 AM  6:25:27 AM  99
33154A36D00F46C0000040A8    6/13/2019   4:54:23 AM  6:10:55 AM  76
33154A36D00F46C000002ECB    6/13/2019   4:55:59 AM  6:10:55 AM  75
33154A36D00F46C000002A2F    6/13/2019   5:03:18 AM  6:20:40 AM  77
33154A36D00F46C000000499    6/13/2019   5:34:35 AM  6:25:27 AM  51
33154A36D00F46C00000627C    6/13/2019   5:38:04 AM  6:25:27 AM  47
33154A36D00F46C000006F74    6/13/2019   5:38:06 AM  6:28:42 AM  50

我现在也在寻找以下内容：

33154A36D00F46C000006F38    6/13/2019              6:28:42 AM   
33154A36D00F46C000006F62    6/13/2019              6:47:42 AM   
33154A36D00F46C000006F90    6/13/2019              7:47:12 AM

Answer 1

您可以做的是创建第二个CTE子句，该子句选择您要描述的内容，并具有相同的字段布局和不相关的NULL。

然后从这两者中进行选择，就完成了

 SELECT * FROM CTE_A 
 UNION ALL 
 SELECT * FROM CTE_B 
          WHERE CTE_B.tag NOT IN (SELECT CTE_A.tag FROM CTE_A)

Answer 2

为什么不简化查询而只做一次通过？

select 
    Tag as 'Tag ID'
    , UID_KEG
    , max(iif(UID_STATION=4,CONVERT(VARCHAR(10), MOVEMENT_DATE, 105),null)) as [DATE]
    , max(iif(UID_STATION=4,CONVERT(VARCHAR(10), MOVEMENT_DATE, 108),null)) as [Time Out]
    , max(iif(UID_STATION=5,CONVERT(VARCHAR(10), MOVEMENT_DATE, 108),null)) as [Time IN]
from MOVEMENT M
Inner join KEG K on 
    K.UNIQUE_ID = M.UID_KEG
    AND Convert(varchar(10),MOVEMENT_DATE,120) = '2019-06-13' 
    and UID_STATION in (4,5)
    and TAG <> 'NO TAG'
group by 
    Tag 
    , UID_KEG

如果您想保留cte的权限，那么首先，由于您要进行内部联接，因此可能会丢失：

where Exists (Select CTE2.[Tag ID] 
       from CTE2
       where CTE2.[Tag ID] = CTE.[Tag ID] )

如果您在[标记ID]上进行内部联接，则只会在两个表中都获得值

我会丢失该where子句，并将其更改为外部联接，在这种情况下，如果希望所有带有in的标签，则将获得所有标签ID（无论它们是入内还是出），或者是左外部联接。日期，像这样：

With CTE AS
( 
select Tag as 'Tag ID', UID_KEG, CONVERT(VARCHAR(10), MOVEMENT_DATE, 105) 
as [DATE],CONVERT(VARCHAR(10), MOVEMENT_DATE, 108) as [Time Out]
from MOVEMENT M
Inner join KEG K on K.UNIQUE_ID = M.UID_KEG
where Convert(varchar(10),MOVEMENT_DATE,120) = '2019-06-13' 
and UID_STATION = 4
and TAG <> 'NO TAG'
) ,
CTE2 AS
(select Tag as 'Tag ID', UID_KEG, CONVERT(VARCHAR(10), MOVEMENT_DATE, 105) 
as [DATE],CONVERT(VARCHAR(10), MOVEMENT_DATE, 108) as [Time IN]
from MOVEMENT M
Inner join KEG K on K.UNIQUE_ID = M.UID_KEG
where Convert(varchar(10),MOVEMENT_DATE,120) = '2019-06-13'
and UID_STATION = 5
and TAG <> 'NO TAG'
)
Select CTE.[Tag ID], CTE.[DATE], [Time IN], [Time Out],DATEDIFF(MINUTE, 
[Time IN], [Time Out]) as [Time in Process]
from CTE
LEFT Join CTE2 on CTE2.[Tag ID] = CTE.[Tag ID]

---已在编辑中添加

如果您希望在出入港时有份量，怎么办：

SELECT
  Tag AS 'Tag ID',
  UID_KEG,
  MAX(iif(UID_STATION = 4, CONVERT(varchar(10), MOVEMENT_DATE, 105), NULL)) AS [DATE],
  MAX(iif(UID_STATION = 4, CONVERT(varchar(10), MOVEMENT_DATE, 108), NULL)) AS [Time Out],
  MAX(iif(UID_STATION = 5, CONVERT(varchar(10), MOVEMENT_DATE, 108), NULL)) AS [Time IN],
  DATEDIFF(MINUTE,
    MAX(iif(UID_STATION = 4, MOVEMENT_DATE, NULL)),
    MAX(iif(UID_STATION = 5, MOVEMENT_DATE, NULL))) AS [Time Elapsed]
FROM MOVEMENT M
INNER JOIN KEG K
  ON K.UNIQUE_ID = M.UID_KEG
  AND CONVERT(varchar(10), MOVEMENT_DATE, 120) = '2019-06-13'
  AND UID_STATION IN (4, 5)
  AND TAG <> 'NO TAG'
GROUP BY Tag,
         UID_KEG

联合结果集

2 个答案: