高分求解sql，也可以使用proc或view只要能解决就行！有点难度，请高手指点！

[showmetoyou]

生成数据表（moblieuser）的脚本如下：
if exists (select * from dbo.sysobjects where id = object_id(N'[dbo].[moblieuser]') and OBJECTPROPERTY(id, N'IsUserTable') = 1)
drop table [dbo].[moblieuser]
GO

CREATE TABLE [dbo].[moblieuser] (
[id] [float] NULL ,
[moblie] [nvarchar] (255) COLLATE Chinese_PRC_CI_AS NULL ,
[password] [nvarchar] (255) COLLATE Chinese_PRC_CI_AS NULL ,
[ordercode] [nvarchar] (255) COLLATE Chinese_PRC_CI_AS NULL ,
[paytime] [smalldatetime] NULL ,
[status] [nvarchar] (255) COLLATE Chinese_PRC_CI_AS NULL ,
[result] [nvarchar] (255) COLLATE Chinese_PRC_CI_AS NULL
) ON [PRIMARY]
GO
数据如下：（精简）
id moblie password ordercode paytime status result
1990.0 13500057554 699704 MOVIE 20041124 NULL 000000
1993.0 13500057554 699704 MOVIE 20041124 1 null
1995.0 13500057554 699704 MOVIE 20041124 1 000000
2051.0 13500337201 741611 MUSIC 20041124 NULL null
2067.0 13500337201 741611 MUSIC 20041124 NULL 000000
2030.0 13500337201 741611 MUSIC 20041124 1 000000
5735.0 13500337201 191406 MUSIC 20041205 NULL 000000
7268.0 13455760459 754181 MOVIE 20041210 NULL 000000
7284.0 13455760459 NULL MOVIE 20041210 1 null
7284.0 13455760459 NULL MUSIC 20041210 1 000000


条件：ordercode字段表示栏目代码， MOVIE为电影频道，订阅收费10元，music为音乐频道，订阅收费5元
status字段为状态标志，只有两个状态，null表示订阅，1表示退订
result字段为处理标志，只有两个状态，000000表示处理成功，null表示处理失败
举例：用户13500057554在20041124日订阅（status为null（is not 1））MOVIE成功（result为000000），20041200对应的收入加10元，
如果不成功，不加钱，如果用户退订MOVICE并且成功，收入减10元，退订不成功，收入不减！
目标：统计生成moblieuser表中三天时间内每天收入表（income）
paytime money
20041124 0 注释0＝（10－0－10＋0＋5－5）
20041205 5
20041210 5 注释5＝（10－0－5）

请各位大虾看看测试，测试，贴出解决方案！分数不够还可以加！
如果有不清楚的地方，直接加我msn：[e-mail]或QQ：39197106！也欢迎加我交流！

[lsxaa]

select a.paytime,
sum(case when a.result is null then 0
else (case when a.status is null then b.val
when a.status=1 then -b.val
end
)
end) as money
from t a,(select 'MOVIE' as ordercode,10 as val union all
'MUSIC' as ordercode,5 as val ) b
where a.ordercode=b.ordercode
group by a.paytime
order by a.paytime

[lsxaa]

我没测试，你自己测试吧

[lsxaa]

select a.paytime,
sum(case when a.result is null then 0
else (case when a.status is null then b.val
when a.status=1 then -b.val
end
)
end) as money
from t a,(select 'MOVIE' as ordercode,10 as val union all
select 'MUSIC' as ordercode,5 as val ) b --改一点
where a.ordercode=b.ordercode
group by a.paytime
order by a.paytime

[子陌红尘]

select
paytime,
sum((case ordercode when 'MOVIE' then 10 when 'MUSIC' then 5 end) * (case when status is null then 1 else -1 end) * (case when result is null then 0 else 1 end)) as money
from
moblieuser
group by
paytime
order by
paytime

[playyuer]

select paytime,
(select sum(b.val) from moblieuser where status is null and result = '0' and paytime = a.paytime)
-
(select sum(b.val) from moblieuser where status = '1' and result = '0' and paytime = a.paytime)

from moblieuser a
,(select 'MOVIE' as ordercode,10 as val
union all
select 'MUSIC',5) b
where a.ordercode = b.ordercode
group by paytime
order by paytime

[vinsonshen]

这样试试吧：

select convert(varchar(10),paytime,120) as paytime,
income=sum(case when ordercode='MOVIE' and status is null and result='000000' then 10
when ordercode='MUSIC' and status is null and result='000000' then 5
when ordercode='MOVIE' and status =1 and result='000000' then -10
when ordercode='MUSIC' and status =1 and result='000000' then -5
else 0 end
)
from moblieuser
group by convert(varchar(10),paytime,120)

[showmetoyou]

谢谢！各位大虾！我测试一下！明天中午揭帖！等着加分吧！

[showmetoyou]

select datename(yy,paytime)+right(('0'+datename(mm,paytime)),2)+right(('0'+datename(dd,paytime)),2) as 日期,
count(distinct moblie) as 用户数目,
sum((case ordercode when 'MSNR001' then 12 when '10057001' then 15 when 'LWZ' then 5 end)
* (case when status is null then 1 else -1 end)
* (case when result is null then 0 else 1 end)) as 总金额
from moblieuser
group by datename(yy,paytime)+right(('0'+datename(mm,paytime)),2)+right(('0'+datename(dd,paytime)),2)
order by datename(yy,paytime)+right(('0'+datename(mm,paytime)),2)+right(('0'+datename(dd,paytime)),2) DESC

这就是我项目中的初步sql，已成功！我明天再研究一下！中午揭帖！

[showmetoyou]

欢迎交流！