Question

我有一个TempTable，其中包含表设置Empname SwipeTime和说明，如下所示

Mike 2019-05-17 12:00:16.383 User Granted Exit From Door Using Reading Device 
Mike 2019-05-17 12:36:11.753 User Granted Entry To Door Using Reading Device
John 2019-05-17 12:00:16.383 User Granted Exit From Door Using Reading Device 
John 2019-05-17 12:36:11.753 User Granted Entry To Door Using Reading Device
Steve 2019-05-17 12:00:16.383 User Granted Exit From Door Using Reading Device 
Steve 2019-05-17 12:36:11.753 User Granted Entry To Door Using Reading Device

在所有人都有不同的滑动时间和描述的情况下，我如何才能获取每位员工的最新记录。

SELECT Distinct
    MAX(EmployeeName) AS EmployeeName,
    MAX(SwipeTime) AS MaxSwipeTime,
    Description
FROM #WhosInOut
GROUP BY EmployeeName,Description, SwipeTime
ORDER BY  EmployeeName, MAX(SwipeTime) ;

我希望列出所有具有最新记录的用户，而不是同时包含退出或进入记录

Answer 1

我认为您想要row_number()：

select t.*
from (select t.*, row_number() over (partition by employeename order by swipetime desc) as seq
      from #WhosInOut t
     ) t
where seq = 1;

Answer 2

您可以执行子查询：

SELECT a.*, b.Description
FROM
    (SELECT Distinct
        EmployeeName EmployeeName,
        MAX(SwipeTime) AS MaxSwipeTime  
     FROM #WhosInOut
     GROUP BY 1)a
JOIN #WhosInOut b
 ON a.EmployeeName = b.EmployeeName and a.MaxSwipeTime = b.SwipeTime
ORDER by EmployeeName, MaxSwipeTime

Answer 3

（可选）相关的子查询将起作用：

SELECT EmployeeName, SwipeTime, Description
FROM #WhosInOut t1
WHERE SwipeTime = (SELECT MAX(SwipeTime) FROM #WhosInOut WHERE t1.EmployeeName = EmployeeName)

Answer 4

您可以使用row_number函数来提取最新的数据：

select x.empname, x.swipetime, x.description
from (
select a.empname, a.swipetime, a.description, ROW_NUMBER() (over partition by a.empname order by a.swipetime desc) as Ranks
from #whosinout a) x
where ranks = 1

我如何为所有用户获得第二行

4 个答案: