近期在做一个考勤系统，考勤主要关注的是缺勤、迟到和早退。眼下的打卡控制器能够记录username和打卡时间，用户可能一天打卡多次，也可能一天仅仅打了一次卡，这些情况都须要考虑。
打卡信息都存储在考勤表中，从中要挖掘出一个月内的缺勤人员，迟到人员和早退人员，而且能显示缺勤、迟到和早退的时间。
考勤表
CREATE TABLE [dbo].[kaoqin]( [user_name] [varchar](50) NULL, [card_time] [datetime] NULL ) ON [PRIMARY] GO
插入測试数据
INSERT INTO [master].[dbo].[kaoqin]
select '张三', '2014-08-03 09:36:00'
union all
select '张三', '2014-08-03 18:10:00'
union all
select '张三', '2014-08-04 08:32:00'
union all
select '张三', '2014-08-04 15:15:00'
union all
select '张三', '2014-08-05 09:32:00'
union all
select '张三', '2014-08-05 15:15:00'
union all
select '张三', '2014-08-01 08:36:00'
union all
select '张三', '2014-08-01 18:10:00'
union all
select '张三', '2014-08-02 08:32:00'
union all
select '张三', '2014-08-02 18:15:00'
union all
select '张三', '2014-08-25 08:00:00'
union all
select '张三', '2014-08-24 19:00:00'
union all
select '张三', '2014-08-27 08:00:00'
union all
select '张三', '2014-08-27 17:00:00'
union all
select '张三', '2014-08-26 10:00:00'
union all
select '张三', '2014-08-26 18:30:00'
union all
select '张三', '2014-08-26 8:00:00'
union all
select '张三', '2014-08-27 18:56:00'
GO
我的思路是用一张暂时表得到这个月的全部工作日。将该暂时表与用户进行交叉连接。这样每一个用户在这个月的每一个工作日都有一条记录。
如果早上9点为上班时间，18点为下班时间，这个能够兴许做成变量的形式。
declare @time_start datetime
declare @time_end datetime
set @time_start = '2014-08-01 00:00:00'
set @time_end = DATEADD(M,1,@time_start)
-- 一个月的工作日
IF object_id('tempdb..#tempDate') is not null
BEGIN
drop table #tempDate
END
CREATE table #tempDate
(
stat_day varchar(10)
)
IF object_id('tempdb..#tempUserDate') is not null
BEGIN
drop table #tempUserDate
END
CREATE table #tempUserDate
(
stat_day varchar(10),
[user_name] varchar(40)
)
CREATE clustered index tempUserDate_Index1 on #tempUserDate ([user_name],stat_day)
declare @time_temp datetime
set @time_temp = @time_start
while @time_temp < @time_end
begin
if datepart(weekday,@time_temp)>1 and datepart(weekday,@time_temp)<7
begin
insert into #tempDate (stat_day) values (CONVERT(varchar(10),@time_temp,121))
end
set @time_temp= dateadd(d,1,@time_temp)
end
insert into #tempUserDate
select * from #tempDate cross join
(select distinct [user_name] from [kaoqin]) t
从原始的kaoqin表中查询出每一个用户的上班时间和下班时间。假设用户一天的打开记录超过两条。那么就会取最早和最晚的一条分别作为上班时间和下班时间。
select [user_name],CONVERT(varchar(10),card_time,121) as stat_day,
MIN(card_time) as on_time,MAX(card_time) as off_time from [kaoqin]
group by [user_name],CONVERT(varchar(10),card_time,121)通过暂时表#tempUserDate和上面的查询结果关联，假设左联接为空，则证明该人员缺勤。
--缺勤
select * from #tempUserDate a
left join
(
select [user_name],CONVERT(varchar(10),card_time,121) as stat_day,
MIN(card_time) as on_time,MAX(card_time) as off_time from [kaoqin]
group by [user_name],CONVERT(varchar(10),card_time,121)
) b on a.[user_name]=b.[user_name] and a.stat_day=b.stat_day
where [b].[user_name] is null以下是迟到和早退的实现SQL。
--迟到
select * from #tempUserDate a
left join
(
select [user_name],CONVERT(varchar(10),card_time,121) as stat_day,
MIN(card_time) as on_time,MAX(card_time) as off_time from [kaoqin]
group by [user_name],CONVERT(varchar(10),card_time,121)
) b on a.[user_name]=b.[user_name] and a.stat_day=b.stat_day
where CONVERT(varchar(100), [b].[on_time], 8)>'09:00:00'
--早退
select * from #tempUserDate a
left join
(
select [user_name],CONVERT(varchar(10),card_time,121) as stat_day,
MIN(card_time) as on_time,MAX(card_time) as off_time from [kaoqin]
group by [user_name],CONVERT(varchar(10),card_time,121)
) b on a.[user_name]=b.[user_name] and a.stat_day=b.stat_day
where CONVERT(varchar(100), [b].[off_time], 8)<'18:00:00'得到的结果
假设某个人他今天既迟到又早退在终于的结果中都会体现。能够从2014-08-05这条数据看出。当然，这个考勤系统还不完好，比如没有将节日考虑进来，初步的考虑是採用Job定期存储每年的节日，假设员工请假，也须要纳入到系统的考虑中。