sql-server - 计算 hh 中的时差 :mm

Question

我有这个查询，但我不确定要使用什么函数，所以我可以获得这种“hh:mm”格式的时差。有人可以帮忙吗？

SELECT        
EI.[FirstName]+' '+EI.[LastName] [EmployeeName], [Dpt].[FullName] [Department], [Desig].[FullName] [Designation],
FirstIN    = CAST(MIN([AttendanceTimeIn]) AS TIME), 
LastOUT    = CAST(MAX([AttendanceTimeOut]) AS TIME), 
HoursSpent = DATEDIFF(HOUR, CAST(MIN(AttendanceTimeIn) AS TIME), CAST(MAX(AttendanceTimeOut) AS TIME)),
CAST(COALESCE(AttendanceTimeIn, AttendanceTimeOut) AS DATE) [Date]

FROM [HRM].[tbl_Designation] [Desig], [HRM].[tbl_Department] [Dpt], [HRM].[tbl_EmployeeInfo] [EI]

FULL OUTER JOIN [HRM].[tbl_EmployeeAttendance] [Attendance] ON [Attendance].[EmpCode] = [EI].[ID] 

WHERE 

[Dpt].[ID] = [EI].[DeptCode] AND [Desig].[ID] = [EI].[DesignationCode]
AND
[EI].[RecordStatusCode] != '13'
AND 
CAST((GETDATE()-1) AS DATE) = CAST(ISNULL([AttendanceTimeIn], [AttendanceTimeOut]) AS Date)

GROUP BY
EI.[FirstName]+' '+EI.[LastName], [Dpt].[FullName], [Desig].[FullName], CAST(COALESCE(AttendanceTimeIn, AttendanceTimeOut) AS DATE)

Answer 1

也许你可以使用这样的东西:

declare @d1 datetime = '2018-01-11 23:40:18.010'
declare @d2 datetime = '2018-01-12 11:59:18.010'

SELECT CONVERT(VARCHAR(2),DATEPART(HOUR,@d2 - @d1)) + ':' + CONVERT(VARCHAR(2),DATEPART(MINUTE,@d2 - @d1));