如何获得日期和下一个日期值的最小值>至少来自sql server [英] How can I get the minimum value of a date and the next date value > minimum from sql server

问题描述

我正在尝试在数据库中显示日期的最小值，只显示下一个值>最低日期值：我的要求：

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

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) 

但没有结果

but no result

推荐答案

为什么不

Why not

SELECT TOP 2 FROM ... ORDER BY DateColumn ASC

多么丑陋的查询！ 其中语句使用了两次以及许多小错误，例如错误使用数字数据类型的撇号！

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

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)

以上查询应该返回正确的数据：最小的 TimeIn 和第二个最小值。

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