以下查询会返回ID列表和时间列表。它从Left表中返回multimple,因为它在右表中显示为多个。
目前它带回了这个:
我希望它从右表中为左表中的每个ID带回最小值。
任何帮助表示赞赏!
set language british
SELECT dbo.Employee.EmployeeID, SUBSTRING(CONVERT(varchar, ClockTemp.ClockTime, 108), 1, 5) as ClockTime
FROM dbo.Employee LEFT JOIN
(
Select [ClockID]
,[EmployeeID]
,[ClockTypeID]
,[ClockDate]
,[ClockTime]
FROM dbo.Clock
WHERE dbo.Clock.ClockDate = DATEADD(DD, 0, CAST('07-11-2016' AS DATE)) AND Clock.ClockTypeID=1
) As ClockTemp
ON dbo.Employee.EmployeeID = ClockTemp.EmployeeID
ORDER BY dbo.Employee.EmployeeID,clocktemp.ClockTime
答案 0 :(得分:0)
也许这个查询对你有用。您不需要按Clocktime订购,因为最小值将从MIN聚合中获得。
如果您使用SQL 2008,我还会将Clocktime字段更改为TIME,以便您可以避免子字符串转换。
set language british
SELECT dbo.Employee.EmployeeID, SUBSTRING(CONVERT(varchar, MIN(ClockTemp.ClockTime), 108), 1, 5) as ClockTime
FROM dbo.Employee LEFT JOIN
(
Select [ClockID]
,[EmployeeID]
,[ClockTypeID]
,[ClockDate]
,[ClockTime]
FROM dbo.Clock
WHERE dbo.Clock.ClockDate = DATEADD(DD, 0, CAST('07-11-2016' AS DATE)) AND Clock.ClockTypeID=1
) As ClockTemp
ON dbo.Employee.EmployeeID = ClockTemp.EmployeeID
GROUP BY dbo.Employee.EmployeeID
ORDER BY dbo.Employee.EmployeeID
答案 1 :(得分:0)
试试:
WITH minClockTimeCTE AS (
SELECT EmployeeID, min(ClockTime) as ClockTime
FROM ClockTemp
GROUP BY EmployeeID
)
SELECT Employee.EmployeeID, minClockTimeCTE.ClockTime
FROM Employee
LEFT OUTER JOIN minClockTimeCTE
ON Employee.EmployeeID = minClockTimeCTE.EmployeeID
答案 2 :(得分:0)
您可以将左连接更改为外部应用顶部1:
SELECT dbo.Employee.EmployeeID,
SUBSTRING(CONVERT(varchar, ClockTemp.ClockTime, 108), 1, 5) as ClockTime
FROM dbo.Employee
OUTER APPLY
(
SELECT TOP 1 [ClockID]
,[EmployeeID]
,[ClockTypeID]
,[ClockDate]
,[ClockTime]
FROM dbo.Clock
WHERE dbo.Clock.ClockDate = DATEADD(DD, 0, CAST('07-11-2016' AS DATE))
AND Clock.ClockTypeID=1
AND dbo.Employee.EmployeeID = ClockTemp.EmployeeID
ORDER BY ClockTime
) As ClockTemp
ORDER BY dbo.Employee.EmployeeID,clocktemp.ClockTime