Question

在下表中，我需要找出登录日期以及用户会话的跨越时间。例如，empid = 1在1/1/2009 @ 3:20登录（typeid = 1）并在同一天3:25注销（typeid = 2）。输出应为：

logged on,      session length,  sessionid
1/1/2009 3:20   5:00             014A3B2C-3874-4D84-A5B3-F234AA2AE30D

每个sessionid都是唯一的。如何创建此类查询？

create table a (empid int, typeid int, created datetime, sessionid nvarchar(36))

insert into a(empid, typeid, created, sessionid) 
values(1, 1, '01/01/2009 3:20:00', '014A3B2C-3874-4D84-A5B3-F234AA2AE30D')
insert into a(empid, typeid, created, sessionid) 
values(1, 2, '01/01/2009 3:25:00', '014A3B2C-3874-4D84-A5B3-F234AA2AE30D')
insert into a(empid, typeid, created, sessionid) 
values(2, 1, '01/09/2009 4:20:00', '014A3B2C-3874-4D84-A5B3-F234AA2AE30E')
insert into a(empid, typeid, created, sessionid) 
values(3, 1, '01/10/2009 3:25:00', '014A3B2C-3874-4D84-A5B3-F234AA2AE30F')
insert into a(empid, typeid, created, sessionid) 
values(2, 2, '01/09/2009 4:23:00', '014A3B2C-3874-4D84-A5B3-F234AA2AE30E')
insert into a(empid, typeid, created, sessionid) 
values(3, 2, '01/10/2009 4:25:00', '014A3B2C-3874-4D84-A5B3-F234AA2AE30F')
select * from a

drop table a

- 编辑 - 有些情况我想过滤掉。例如，对于empid = 1，可能是这样：

insert into a(empid, typeid, created, sessionid) 
values(1, 1, '01/01/2009 3:20:00', '014A3B2C-3874-4D84-A5B3-F234AA2AE30D')
insert into a(empid, typeid, created, sessionid) 
values(1, 1, '01/01/2009 3:25:00', '014A3B2C-3874-4D84-A5B3-F234AA2AE30D')

上面的empid没有匹配的typeid = 2。有两个typeid = 1。

Answer 1

select empid,sessionid,
    min(created),max(created),
    datediff(mi,min(created),max(created)) 
from a 
group by empid,sessionid

Answer 2

SELECT LogOn.[created] as [logged on], DATEDIFF(day,LogOn.[created],LogOff.[created]) + ' ' + DATEDIFF(hour,LogOn.[created],LogOff.[created]) + ':' + DATEDIFF(minutes,LogOn.[created],LogOff.[created]) + ':' + DATEDIFF(second,LogOn.[created],LogOff.[created]), a.[sessionid] 
FROM a as [LogOn] 
JOIN a as [LogOff] ON LogOn.[sessionid] = LogOff.[sessionid] 
WHERE LogOn.[typeid] = 1 AND LogOff.[typeid] = 2

Answer 3

SELECT  SessionId,
        MIN(CASE WHEN typeid = 1 THEN created END) [logged on],
        DATEDIFF(SECOND,MIN(CASE WHEN typeid = 1 THEN created END),MIN(CASE WHEN typeid = 2 THEN created END)) [session length]
FROM a
GROUP BY sessionId

但是，您需要操纵会话长度以获得所需的输出格式。

按一列和两行分组

3 个答案: