日期:2014-05-18  浏览次数:20860 次

有点难的查询,请求帮忙?《有点难的查询,请求帮忙?〉
销售表x_sale
sale_id sale_addtime sale_money
  1 2012-01-01 500
  2 2012-01-01 500
  3 2012-01-02 300

销售退货表x_saleout
saleout_id saleout_addtime saleout_money
  1 2012-01-01 100
  1 2012-01-01 100
  2 2012-01-03 200

如何才能得到
addtime sale_money saleout_money money
2012-01-01 1000 200 800
2012-01-02 300 0 300
2012-01-03 0 200 -200

就是同时得到两个表的数据,并且带子查询的。

------解决方案--------------------
SQL code

--> 测试数据:[x_sale]
if object_id('[x_sale]') is not null drop table [x_sale]
create table [x_sale]([sale_id] int,[sale_addtime] datetime,[sale_money] int)
insert [x_sale]
select 1,'2012-01-01',500 union all
select 2,'2012-01-01',500 union all
select 3,'2012-01-02',300
--> 测试数据:[x_saleout]
if object_id('[x_saleout]') is not null drop table [x_saleout]
create table [x_saleout]([saleout_id] int,[saleout_addtime] datetime,[saleout_money] int)
insert [x_saleout]
select 1,'2012-01-01',100 union all
select 1,'2012-01-01',100 union all
select 2,'2012-01-03',200


with t
as(
select 
    [sale_addtime],
    SUM([sale_money]) [sale_money]
from 
    [x_sale]
group by 
    [sale_addtime]
),
m as(
select 
    [saleout_addtime],
    SUM([saleout_money]) as [saleout_money]
from 
    [x_saleout]
group by
    [saleout_addtime]
)
select 
    isnull(t.sale_addtime,m.saleout_addtime) as addtime,
    isnull(t.sale_money,0) as sale_money,
    isnull(m.saleout_money,0) as saleout_money,
    isnull(t.sale_money,0)-isnull(m.saleout_money,0) as [money]
from 
    t 
full join 
    m 
on 
    t.sale_addtime=m.saleout_addtime
/*
addtime    sale_money    saleout_money    money
---------------------------------------------------
2012-01-01 00:00:00.000    1000    200    800
2012-01-02 00:00:00.000    300    0    300
2012-01-03 00:00:00.000    0    200    -200
*/