Question

我的考勤表中有以下数据：

Staff_ID  |       User_Attend_Date    |  Business_Day 
  S01     |   2013-05-01 15:18:45.537 |   2013-05-01  
  S01     |   2013-05-02 00:00:00.000 |   2013-05-01  
  S02     |   2013-05-03 06:20:30.225 |   2013-05-02

我想显示一个新列，检查员工入住日期减3小时是否与营业日期相同。与营业日相同，保留User_Attend_Date.My_Day的数据类型为DATE，其他2列为DATETIME。结果如下：

 Staff_ID  |       User_Attend_Date    |  Business_Day   |     Check_Same_Day
   S01     |   2013-05-01 15:18:45.537 |   2013-05-01    | 2013-05-01 15:18:45.537
   S01     |   2013-05-02 00:00:00.000 |   2013-05-01    | 2013-05-01 15:18:45.537
   S02     |   2013-05-03 06:20:30.225 |   2013-05-02    | 2013-05-03 06:20:30.225

我尝试了以下代码，但未能达到结果。

SELECT *, Case WHEN DATEADD(hh, -3, User_Attend_Date)=Business_Day 
THEN User_Attend_Date ELSE BusinessDay END As 'Check_Same_Day' FROM tblAttendance

我尝试使用分区但也失败了。

DATEADD(hh, -3, User_Attend_Date) OVER (PARTITION BY BusinessDay) AS 'Check_Same_Day'

希望有人可以教我。谢谢

Answer 1

要将日期时间更改为其日期，请使用：

SELECT DATEADD(d, DATEDIFF(d, 0, GETDATE()), 0), GETDATE()

所以在你的情况下，你可以用它来与Business_Day

进行比较

SELECT Case WHEN DATEADD(d, DATEDIFF(d, 0, DATEADD(hour, -3, User_Attend_Date)), 0) = Business_Day

修改

您实际上想要为用户使用任何给定日期的第一次登录，因此您可以使用CTE来实现此目的。我添加了一个“期望”列（您可以轻松删除），我假设您必须有一个user_id来识别用户？如果是，请尝试：

http://sqlfiddle.com/#!3/c54b4/1

WITH AttendanceDate AS ( SELECT User_Id, User_Attend_Date, Business_Day, DATEADD(d, DATEDIFF(d, 0, DATEADD(hour, -3, User_Attend_Date)), 0) AS ActualAttendDate FROM tblAttendance ), FirstAttendance AS ( SELECT User_Id, ActualAttendDate, MIN(User_Attend_Date) FirstLogin FROM AttendanceDate GROUP BY User_Id, ActualAttendDate ) SELECT TA.User_Id, TA.User_Attend_Date, TA.Business_Day, FA.FirstLogin FROM AttendanceDate AS TA INNER JOIN FirstAttendance AS FA ON TA.User_Id = FA.User_Id AND TA.ActualAttendDate = FA.ActualAttendDate

请注意，演员阵容现在有点多余：http://sqlfiddle.com/#!3/c54b4/2，以及多个用户看到它：http://sqlfiddle.com/#!3/4f2c7/1。没有Expect作为列的最终结果位于http://sqlfiddle.com/#!3/e9e6c/1。

编辑2

如果您不需要User_Id分组，那么它就像（http://sqlfiddle.com/#!3/0daab/1）一样简单：

WITH AttendanceDate AS ( SELECT User_Attend_Date, Business_Day, DATEADD(d, DATEDIFF(d, 0, DATEADD(hour, -3, User_Attend_Date)), 0) AS ActualAttendDate FROM tblAttendance ), FirstAttendance AS ( SELECT ActualAttendDate, MIN(User_Attend_Date) FirstLogin FROM AttendanceDate GROUP BY ActualAttendDate ) SELECT TA.User_Attend_Date, TA.Business_Day, FA.FirstLogin FROM AttendanceDate AS TA INNER JOIN FirstAttendance AS FA ON TA.ActualAttendDate = FA.ActualAttendDate

原始答案（根据解释预期结果如何组合的新信息不再相关）

完整（http://sqlfiddle.com/#!3/fa5bb/10）：

SELECT User_Attend_Date, DATEADD(hour, -3, User_Attend_Date), DATEADD(d, DATEDIFF(d, 0, DATEADD(hour, -3, User_Attend_Date)), 0) AS 'ForClarity1', CAST(DATEADD(d, DATEDIFF(d, 0, DATEADD(hour, -3, User_Attend_Date)), 0) AS DATE) AS 'ForClarity2', Business_Day AS 'ForClarity3', Case WHEN CAST(DATEADD(d, DATEDIFF(d, 0, DATEADD(hour, -3, User_Attend_Date)), 0) AS DATE) = Business_Day THEN User_Attend_Date ELSE Business_Day END As 'Check_Same_Day' FROM dbo.tblAttendance

Answer 2

我看到两个选项：

1）DateDiff：

SELECT Case WHEN datediff(day,DATEADD(hh, -3, User_Attend_Date),Business_Day) < 1
THEN User_Attend_Date ELSE BusinessDay END As 'Check_Same_Day' FROM tblAttendance

2）演员：

SELECT Case WHEN Cast(DATEADD(hh, -3, User_Attend_Date) as Date) = Business_Day
THEN User_Attend_Date ELSE BusinessDay END As 'Check_Same_Day' FROM tblAttendance

Answer 3

你几乎就在那里，问题是你需要将DATEADD的结果转换为DATE（而不是DATETIME）：

DECLARE @att TABLE
(
    User_Attend_Date DATETIME,
    Business_Day DATE
)

INSERT INTO @att
SELECT '2013-05-01 15:18:45.537', '2013-05-01'
UNION ALL
SELECT '2013-05-02 00:00:00.000', '2013-05-01'
UNION ALL
SELECT '2013-05-03 06:20:30.225', '2013-05-02'

SELECT 
    User_Attend_Date,
    Business_Day,
    CASE 
        WHEN CAST(DATEADD(hh, -3, User_Attend_Date) AS DATE) = Business_Day THEN User_Attend_Date
        ELSE Business_Day
    END AS 'Check_Same_Day'
FROM @att

这会给出您想要的结果，表明其中两个项目符合您的规则。

Answer 4

您需要将第一列（日期时间）转换为日期以与日期进行比较。

 SELECT 
      *,
      CASE WHEN  CAST(DATEADD(HOUR,-4,User_Attend_Date) AS DATE)=Business_Day
        THEN User_Attend_Date 
        ELSE Business_Day 
      END AS Check_Same_Day
      FROM #test

Answer 5

我认为它在行中显示秒值，因为你有Bussiness_day的数据类型DATE和其他列的数据类型DATETIME。

那么可以在sql 2008中使用DATE数据类型吗？

Answer 6

SELECT  USER_ATTEND_DATE,SUBSTRING(CAST(BUSINESS_DAY AS VARCHAR(20)),0,13)AS BUSINESS_DAY,

        CASE 
            WHEN  DATEADD(hh, -3, User_Attend_Date) = Business_Day 
            THEN Business_Day
            ELSE DATEADD(hh, -3, User_Attend_Date)
        END CHECK_SAME_DAY
FROM attendance

查看当天的表格

6 个答案: