Sql 如何匹配这些值以进行比较
我需要得到两行中两个时间的日期差。我试图操纵此表的ID,但发现了一个死胡同 该表如下所示:Sql 如何匹配这些值以进行比较,sql,sql-server,datediff,Sql,Sql Server,Datediff,我需要得到两行中两个时间的日期差。我试图操纵此表的ID,但发现了一个死胡同 该表如下所示: ID TimeIn TimeOut EmpId 1 8.30am 12.30pm usr1 2 1.30pm 5.30pm usr1 我需要得到第一行的超时和第二行的Timein之间的时差 预期输出应如下所示: ID TimeIn TimeOut EmpId 1
ID TimeIn TimeOut EmpId
1 8.30am 12.30pm usr1
2 1.30pm 5.30pm usr1
我需要得到第一行的超时和第二行的Timein之间的时差
预期输出应如下所示:
ID TimeIn TimeOut EmpId
1 8.30am 12.30pm usr1
2 1.30pm 5.30pm usr1
午餐时间
usr1 1:00
1:00相当于1小时。我想你真正想问的是,如何匹配这些值以进行比较。一旦两个值都有了,比较就很简单了,但是匹配成对的行就不那么简单了。查看这篇文章,了解如何一步一步地解决类似问题:以下是我的代码,并在MYSQL中进行测试,以供参考
//EMP table content like below
id TimeIn TimeOut EmpId
1 2015-03-21 09:00:00 2015-03-21 12:30:00 usr1
2 2015-03-21 13:30:00 2015-03-21 17:30:00 usr1
3 2015-03-21 08:00:00 2015-03-21 13:30:00 usr2
4 2015-03-21 14:00:00 2015-03-21 16:00:00 usr2
// find every empId TimeOut of first row and TimeIn of second row
SELECT id,TimeOut ,EmpId,
(select c.TimeIn from EMP c where c.EmpId = a.EmpId and c.id > a.id limit 1) as 'TimeIn'
from EMP a
//result as below
1 2015-03-21 12:30:00 usr1 2015-03-21 13:30:00
2 2015-03-21 17:30:00 usr1 NULL
3 2015-03-21 13:30:00 usr2 2015-03-21 14:00:00
4 2015-03-21 16:00:00 usr2 NULL
//find the timediff between TimeIn and TimeOut
select id,EmpId,timediff(TimeIn,TimeOut) as diff from
(
SELECT id,TimeOut ,EmpId,
(select c.TimeIn from EMP c where c.EmpId = a.EmpId and c.id > a.id limit 1) as 'TimeIn'
from EMP a
) as tmp
// result as below
id EmpId diff
1 usr1 01:00:00
2 usr1 NULL
3 usr2 00:30:00
4 usr2 NULL
希望这能对您有所帮助。根据您的数据库是否支持,请使用LAG函数。@Randy,LAG函数?从没听说过。我目前正在使用Sql Server。到目前为止您尝试了什么?TimeIn和TimeOut的数据类型是什么?@KhurramAli,我原来的查询比这个复杂一点。我已经尝试使用嵌套选择功能来区分ID。