How can I get the minimum value of a date and the next date value > minimum from sql server

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SQL-Server-2008R2

I'm trying to display the minimum valus of a date in my database and just the next value > minimum date value: my request:

SQL

select  min(Times_IN) as MinTimeIn  ,
(select min(Times_IN) from Emp_Attendance att where att.Times_IN> min (Times_IN) and month (Times_IN) ='4' and Emp_Attendance.EmpID='31'),
 max (Times_IN) as MaxTimeIn, min(Times_Out) as MinTimeOut, max(Times_Out) as MaxTimeOut,CAST(_Date AS DATE) as DateAttendance from Emp_Attendance where month (Times_IN) ='4' and Emp_Attendance.EmpID='31' group by CAST(_Date AS DATE)

but no result

Posted 8-Jun-15 8:12am

Leila Toumi

Updated 8-Jun-15 8:16am

v2

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OriginalGriff · Answer 1 · 2015-06-08T08:31:00

Solution 1

Why not

SQL

SELECT TOP 2 FROM ... ORDER BY DateColumn ASC

Posted 8-Jun-15 8:31am

OriginalGriff

Comments

Leila Toumi 8-Jun-15 14:40pm

I tried it but no result

Maciej Los · Answer 2 · 2015-06-08T11:49:00

What an ugly query! Where statement is used twice and many minor errors, such as incorrect usage of apostrophe with numeric data type!

SQL

SELECT MinTimeIn 
FROM (
    SELECT Times_IN AS MinTimeIn, ROW_NUMBER() OVER(ORDER BY Times_IN) AS RowNo
    FROM Emp_Attendance att
    WHERE month(Times_IN) = 4 AND EmpID=31
) AS T
WHERE RowNo IN (1,2)

Above query should return proper data: the smallest TimeIn and the second minimum value.