Sql Server中Update用法的心得

Sql Server中Update用法的心得

问题： 表中有两字段：id_no (varchar) , in_date (datetime) ，把in_date 相同的记录的in_date依次累加1秒，

使in_date没有相同的记录。

如以下原始数据:
id_no in_date
5791 2003-9-1 14:42:02
5792 2003-9-1 14:42:02
5794 2003-9-1 14:42:02
5795 2003-9-1 14:42:03
5796 2003-9-1 14:42:03
5797 2003-9-1 14:42:03
5831 2003-9-1 14:42:04
5832 2003-9-1 14:42:14
5833 2003-9-1 14:42:14

得到结果是：

id_no in_date
5791 2003-9-1 14:42:02
5792 2003-9-1 14:42:03
5794 2003-9-1 14:42:04
5795 2003-9-1 14:42:05
5796 2003-9-1 14:42:06
5797 2003-9-1 14:42:07
5831 2003-9-1 14:42:08
5832 2003-9-1 14:42:14
5833 2003-9-1 14:42:15

处理方法：


--建立测试环境


create table a(id_no varchar(8),in_date datetime)


go


insert into a select '5791','2003-9-1 14:42:02'


union all select '5792','2003-9-1 14:42:02'


union all select '5794','2003-9-1 14:42:02'


union all select '5795','2003-9-1 14:42:03'


union all select '5796','2003-9-1 14:42:03'


union all select '5797','2003-9-1 14:42:03'


union all select '5831','2003-9-1 14:42:04'


union all select '5832','2003-9-1 14:42:04'


union all select '5833','2003-9-1 14:42:04'


union all select '5734','2003-9-1 14:42:02'


union all select '6792','2003-9-1 14:42:22'


union all select '6794','2003-9-1 14:42:22'


union all select '6795','2003-9-1 14:42:23'


union all select '6796','2003-9-1 14:42:23'


union all select '6797','2003-9-1 14:42:23'


union all select '6831','2003-9-1 14:42:34'


union all select '6832','2003-9-1 14:42:34'


union all select '6833','2003-9-1 14:42:54'


union all select '6734','2003-9-1 14:42:22'


go


--生成临时表，按照in_date排序


select * into # from a order by in_date


--相同的时间，加一秒。加完了不带重复的


declare @date1 datetime,@date2 datetime,@date datetime


update #


set @date=case when @date1=in_date or @date2>=in_date then dateadd(s,1,@date2) else in_date end,


@date1=in_date,


@date2=@date,


in_date=@date


--更新到基本表中去


update a set a.in_date=b.in_date from a a join # b on a.id_no=b.id_no




select * from a


drop table #,a