sql的大全代码（5）

复杂年月处理.sql

SQL code

<!--

Code highlighting produced by Actipro CodeHighlighter (freeware)
http://www.CodeHighlighter.com/

-->--定义基本数字表
declare @T1 table(代码 int,名称 varchar(10),参加时间 datetime,终止时间 datetime)
insert into @T1
    select 12,'单位1','2003/04/01','2004/05/01'
    union all select 22,'单位2','2001/02/01','2003/02/01'
    union all select 42,'单位3','2000/04/01','2003/05/01'
    union all select 25,'单位5','2003/04/01','2003/05/01'

--定义年表
declare @NB table(代码 int,名称 varchar(10),年份 int)
insert into @NB
    select 12,'单位1',2003
    union all select 12,'单位1',2004
    union all select 22,'单位2',2001
    union all select 22,'单位2',2002
    union all select 22,'单位2',2003

--定义月表
declare @YB table(代码 int,名称 varchar(10),年份 int,月份 varchar(2))
insert into @YB
    select 12,'单位1',2003,'04'
    union all select 22,'单位2',2001,'01'
    union all select 22,'单位2',2001,'12'

--为年表+月表数据处理准备临时表
select top 8246 y=identity(int,1753,1)
into #tby from
    (select id from syscolumns) a,
    (select id from syscolumns) b,
    (select id from syscolumns) c

--为月表数据处理准备临时表
select top 12 m=identity(int,1,1)
into #tbm from syscolumns

/*--数据处理--*/
--年表数据处理
select a.*
from(
select a.代码,a.名称,年份=b.y
from @T1 a,#tby b
where b.y between year(参加时间) and year(终止时间)
) a left join @NB b on a.代码=b.代码 and a.年份=b.年份
where b.代码 is null

--月表数据处理
select a.*
from(
select a.代码,a.名称,年份=b.y,月份=right('00'+cast(c.m as varchar),2)
from @T1 a,#tby b,#tbm c
where b.y*100+c.m between convert(varchar(6),参加时间,112) 
    and convert(varchar(6),终止时间,112)
) a left join @YB b on a.代码=b.代码 and a.年份=b.年份 and a.月份=b.月份
where b.代码 is null
order by a.代码,a.名称,a.年份,a.月份

--删除数据处理临时表
drop table #tby,#tbm

交叉表.sql

SQL code

<!--

Code highlighting produced by Actipro CodeHighlighter (freeware)
http://www.CodeHighlighter.com/

-->--示例

--示例数据
create table tb(ID int,Time datetime)
insert tb select 1,'2005/01/24 16:20'
union all select 2,'2005/01/23 22:45'
union all select 3,'2005/01/23 0:30'
union all select 4,'2005/01/21 4:28'
union all select 5,'2005/01/20 13:22'
union all select 6,'2005/01/19 20:30'
union all select 7,'2005/01/19 18:23'
union all select 8,'2005/01/18 9:14'
union all select 9,'2005/01/18 18:04'
go

--查询处理：
select     case when grouping(b.Time)=1 then 'Total' else b.Time end,
    [Mon]=sum(case a.week when 1 then 1 else 0 end),
    [Tue]=sum(case a.week when 2 then 1 else 0 end),
    [Wed]=sum(case a.week when 3 then 1 else 0 end),
    [Thu]=sum(case a.week when 4 then 1 else 0 end),
    [Fri]=sum(case a.week when 5 then 1 else 0 end),
    [Sat]=sum(case a.week when 6 then 1 else 0 end),
    [Sun]=sum(case a.week when 0 then 1 else 0 end),
    [Total]=count(a.week)
from(
    select Time=convert(char(5),dateadd(hour,-1,Time),108)
            --时间交界点是1am,所以减1小时，避免进行跨天处理
        ,week=(@@datefirst+datepart(weekday,Time)-1)%7
            --考虑@@datefirst对datepart的影响
    from tb
)a right join(
    select id=1,a='16:00',b='19:59',Time='[5pm - 9pm)' union all
    select id=2,a='20:00',b='23:59',Time='[9pm - 1am)' union all
    select id=3,a='00:00',b='02:59',Time='[1am - 4am)' union all
    select id=4,a='03:00',b='07:29',Time='[4am - 8:30am)' union all
    select id=5,a='07:30',b='11:59',Time='[8:30am - 1pm)' union all
    select id=6,a='12:00',b='15:59',Time='[1pm - 5pm)'
)b on a.Time>=b.a and a.Time<b.b
group by b.id,b.Time with rollup
having grouping(b.Time)=0 or grouping(b.id)=1
go

--删除测试
drop table tb

/*--测试结果

               Mon   Tue   Wed   Thu   Fri   Sat   Sun   Total 
-------------- ----- ----- ----- ----- ----- ------ ---- -------
[5pm - 9pm)    0     1     2     0     0     0     0     3
[9pm - 1am)    0     0     0     0     0     0     2     2
[1am - 4am)    0     0     0     0     0     0     0     0
[4am - 8:30am) 0     0     0     0     1     0     0     1
[8:30am - 1pm) 0     1     0     0     0     0     0     1
[1pm - 5pm)    1     0     0     1     0     0     0     2
Total          1     2     2     1     1     0     2     9

（所影响的行数为 7 行）
--*/

任意两个时间之间的星期几的次数-横.sql

SQL code

<!--

Code highlighting produced by Actipro CodeHighlighter (freeware)
http://www.CodeHighlighter.com/

-->if exists (select * from dbo.sysobjects where id = object_id(N'[dbo].[f_weekdaycount]') and xtype in (N'FN', N'IF', N'TF'))
drop function [dbo].[f_weekdaycount]
GO

/*--计算任意两个时间之间的星期几的次数(横向显示)

    本方法直接判断 @@datefirst 做对应处理
    不受 sp_language 及 set datefirst 的影响     

--邹建 2004.08(引用请保留此信息)--*/

/*--调用示例
    
    select * from f_weekdaycount('2004-9-01','2004-9-02')
--*/
create function f_weekdaycount(
@dt_begin datetime,
@dt_end datetime
)returns table
as
return(
    select 跨周数
        ,周一=case a
            when -1 then case when 1 between b and c then 1 else 0 end
            when  0 then case when b<=1 then 1 else 0 end
                    +case when c>=1 then 1 else 0 end
            else a+case when b<=1 then 1 else 0 end
                +case when c>=1 then 1 else 0 end
            end
        ,周二=case a
            when -1 then case when 2 between b and c then 1 else 0 end
            when  0 then case when b<=2 then 1 else 0 end
                    +case when c>=2 then 1 else 0 end
            else a+case when b<=2 then 1 else 0 end
                +case when c>=2 then 1 else 0 end
            end
        ,周三=case a
            when -1 then case when 3 between b and c then 1 else 0 end
            when  0 then case when b<=3 then 1 else 0 end
                    +case when c>=3 then 1 else 0 end
            else a+case when b<=3 then 1 else 0 end
                +case when c>=3 then 1 else 0 end
            end
        ,周四=case a
            when -1 then case when 4 between b and c then 1 else 0 end
            when  0 then case when b<=4 then 1 else 0 end
                    +case when c>=4 then 1 else 0 end
            else a+case when b<=4 then 1 else 0 end
                +case when c>=font-weight: bold; color: #8

  


  

    
    
分享到：
      
      
    
  

  

    

      sql的大全代码（6）
      |
      sql的大全代码（4）
    
  
  

    

      
2009-04-17 12:12
      
浏览 827
      
评论(0)
      
      
            
      
查看更多