Question

我的记录如下

 Login Time                   Logout Time            User   Shift
2017-04-01 21:30:00.000   2017-04-02 00:00:00.000   I0402   N
2017-04-02 00:00:00.000   2017-04-02 00:30:00.000   I0402   N
2017-04-02 01:30:00.000   2017-04-02 05:30:00.000   I0402   N
2017-04-02 06:30:00.000   2017-04-02 08:30:00.000   I0402   N

我希望为用户获得最早的进入时间和最新的注销时间。我怎么能这样做？

我希望记录如下

   Login Time                  Logout Time              User    Shift
2017-04-01 21:30:00.000      2017-04-02 08:30:00.000   I0402    N

有可能吗？感谢。

修改

我之前尝试使用Max和Min，并按用户和Shift进行分组，这是有效的。如果记录超过1天，那似乎不起作用

2017-04-01 21:30:00.000   2017-04-02 00:00:00.000   I0402   N
2017-04-02 00:00:00.000   2017-04-02 00:30:00.000   I0402   N
2017-04-02 01:30:00.000   2017-04-02 05:30:00.000   I0402   N
2017-04-02 06:30:00.000   2017-04-02 08:30:00.000   I0402   N 
2017-04-02 21:30:00.000   2017-04-03 00:00:00.000   I0402   N
2017-04-03 00:00:00.000   2017-04-03 00:30:00.000   I0402   N
2017-04-03 01:30:00.000   2017-04-03 05:30:00.000   I0402   N

预期结果如下

2017-04-01 21:30:00.000      2017-04-02 08:30:00.000   I0402    N
2017-04-02 21:30:00.000      2017-04-03 05:30:00.000   I0402    N

Answer 1

SELECT
 User 
,MIN(Login Time)
,MAX(Logout Time)
,Shift
FROM
   dbo.table
GROUP BY
 User,Shift

Answer 2

我之前尝试使用Max和Min，并按用户和Shift进行分组，这是有效的。如果记录超过1天，那似乎不起作用

2017-04-01 21:30:00.000   2017-04-02 00:00:00.000   I0402   N
2017-04-02 00:00:00.000   2017-04-02 00:30:00.000   I0402   N
2017-04-02 01:30:00.000   2017-04-02 05:30:00.000   I0402   N
2017-04-02 06:30:00.000   2017-04-02 08:30:00.000   I0402   N                     
2017-04-02 21:30:00.000   2017-04-03 00:00:00.000   I0402   N
2017-04-03 00:00:00.000   2017-04-03 00:30:00.000   I0402   N
2017-04-03 01:30:00.000   2017-04-03 05:30:00.000   I0402   N

预期结果如下

2017-04-01 21:30:00.000      2017-04-02 08:30:00.000   I0402    N
2017-04-02 21:30:00.000      2017-04-03 05:30:00.000   I0402    N

Answer 3

您可以使用以下查询来获取结果

 SELECT T1.USER, T2.MAX_LOGIN_TIME, T3.MIN_LOGIN_TIME FROM 
 TABLE_NAME T1
 INNER JOIN (SELECT MAX(LOGIN_TIME) AS MAX_LOGIN_TIME,USER FROM TABLE_NAME 
 GROUP BY USER) T2
 ON (T1.USER = T2.USER)
 INNER JOIN (SELECT MIN(LOGIN_TIME) AS MIN_LOGIN_TIME,USER FROM TABLE_NAME 
 GROUP BY USER) T3
 ON (T1.USER = T3.USER);

Answer 4

看起来你真正想要的是每天一行，用户和班次。所以按这些分组并花费最小和最长时间。

select min(Login_time) as login, max(logout_time) as logout, user, shift
from mytable
group by date(login_time), user, shift
order by date(Login_time), user, shift;

这里DATE(login_time)只是一个例子。如何从日期时间获取日期不同于DBMS到DBMS。查找DBMS的日期/时间函数以获得适当的功能。

选择最大和最小日期时间

4 个答案: