无法从每个后退日期的日期时间列中获取MIN和MAX时间

Question

我正在尝试从MIN（MAX）列获取Datetime和InOut时间，但两列的输出相同。

我的查询：

SELECT
  CONVERT(NVARCHAR(12),MIN(Punch_history.Ecode)) as EmpCode,
  CONVERT(NVARCHAR(12),MIN(EmployeeMaster.RecommandedBy)) as EmpID,
  convert(date, InOut) as Report_date,
  CONVERT(VARCHAR(10),(InOut),108) as InTime,
  CONVERT(VARCHAR(10),(InOut),108) as OutTime,
  CONVERT(NVARCHAR(12),MIN(TID)) as LOCATION
  FROM Punch_history inner join EmployeeMaster on Punch_history.ECode = EmployeeMaster.Ecode
  where CAST (InOut as DATE) between  CAST(getdate() -1 as DATE ) and CAST(getdate() -1 as DATE ) and 
  EmployeeMaster.RecommandedBy like 'M0%' group by EmpID,InOut

Answer 1

尝试按转换日期而不是inout进行分组。

SELECT
  CONVERT(NVARCHAR(12),MIN(Punch_history.Ecode)) as EmpCode,
  convert(date, InOut) as Report_date,
  CONVERT(NVARCHAR(10),min(InOut),108) as InTime,
  CONVERT(NVARCHAR(10),max(InOut),108) as OutTime
  FROM Punch_history 
  where CAST (InOut as DATE) between  CAST(getdate() -1 as DATE ) 
    and CAST(getdate() -1 as DATE ) 
  group by EmpID, convert(date, InOut)

Answer 2

除了分组问题，你的where子句很奇怪。首先使用日期数学的快捷方式是一个没有胜利的情况。它会因datetime2而失败。在两个相同的值之间使用是很奇怪的。

SELECT
    CONVERT(NVARCHAR(12), MIN(Punch_history.Ecode)) as EmpCode,
    convert(date, MIN(InOut)) as Report_date,
    CONVERT(NVARCHAR(10), min(InOut), 108) as InTime,
    CONVERT(NVARCHAR(10), max(InOut), 108) as OutTime
FROM Punch_history 
where convert(date, InOut) = convert(date, dateadd(day, -1, getdate()))
group by EmpID

Answer 3

主要问题是你不应该按InOut（datetime）分组，这就是为什么在结果中max和min是相同的。实际上，这取消了您希望每天进行的分组。

  SELECT
  CONVERT(NVARCHAR(12),MIN(Punch_history.Empcode)) as EmpCode,
  CONVERT(DATE , Max(InOut)) as Report_date,
  CONVERT(NVARCHAR(10),min(InOut),108) as InTime,
  CONVERT(NVARCHAR(10),max(InOut),108) as OutTime
  FROM Punch_history 
  where datediff(day, CAST(InOut as DATE ) , CAST(getdate() as DATE ) )=1
  group  by EmpID

另外，如果empid和empcode是1-1（从逻辑上讲它们应该是）    然后你可以直接用empcode或两者进行分组    避免这种相当奇怪的Min（empcode）。