Question

每天都有船只抵达一个有“P”级乘客的港口。每艘船的名称为“N”，每天最多到达一次。我有一张这样的数据表：

Date          N          P

2013-11-27    A         12
2013-11-27    G         16
2013-11-27    B         56
2013-11-28    A         10
2013-11-29    B         19
2013-11-29    A         19
2013-11-30    G         18
2013-11-30    B         13
2013-11-30    A         11

我希望在给定的时间跨度内将一艘船的'A'乘客数量与另一艘'A'乘客数量进行比较，但仅限于他们在同一天到达时。结果是

Date          PA        PB

2013-11-27    12        56
2013-11-29    19        19
2013-11-30    11        13

注意28日没有'B'，所以既没有报告（反之亦然）。

同桌上的某种外连接？

SELECT A.Date, A.P as PA, B.P AS PB 
FROM Arrivals A OUTER JOIN Arrivals B ON A.Date = B.Date
WHERE ... ???

Date >= @start AND Date <= @end AND
A.N = 'A' AND B.N = 'B'

以上查询是我想到的开始，但我不知道如何完成它。有谁知道查询？ SQL Server 2008 谢谢你的帮助。

Answer 1

这有用吗？

SELECT A.Date, A.P as PA, B.P AS PB 
FROM (Select * FROM Arrivals WHERE N = 'A') A 
INNER JOIN (Select * FROM Arrivals WHERE N = 'B') B 
ON A.Date = B.Date

Answer 2

试试这个：

create table #tmp 
(
dt date, 
n varchar (1) , 
p int
)
GO

insert into #tmp
values 
('2013-11-27'   , 'A'      ,   12),
('2013-11-27' ,   'G'  ,       16),
('2013-11-27'   , 'B'       ,  56),
('2013-11-28' ,   'A' ,        10),
('2013-11-29'   , 'B'  ,       19),
('2013-11-29'   , 'A' ,        19),
('2013-11-30'   , 'G' ,        18),
('2013-11-30'   , 'B' ,        13),
('2013-11-30'   , 'A' ,        11)

GO

select distinct 
a.dt,
(select SUM(p) from #tmp y where y.dt = a.dt and y.n = 'A') pa,
(select SUM(p) from #tmp y where y.dt = a.dt and y.n = 'B') pb,
(select SUM(p) from #tmp y where y.dt = a.dt and y.n = 'G') pG
from #tmp a
group by a.dt, a.n

Go 
drop table #tmp

Answer 3

;WITH ARRIVAL_A AS
(SELECT DATE, P FROM ARRIVALS WHERE N = 'A'),
ARRIVAL_B AS
(SELECT DATE, P FROM ARRIVALS WHERE N = 'B')
SELECT ARRIVAL_A.DATE, ARRIVAL_A.P [PA], ARRIVAL_B.P [PB]
FROM ARRIVAL_A INNER JOIN ARRIVAL_B
ON ARRIVAL_A.DATE = ARRIVAL_B.DATE

加入相同的表匹配日期范围

3 个答案: