关于论坛上那个SQL微软面试题。我的解答方法 :-)

2024-07-21 02:08:57

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问题：

一百个账户各有100$，某个账户某天如有支出则添加一条新记录，记录其余额。一百天后，请输出每天所有账户的余额信息

这个问题的难点在于每个用户在某天可能有多条纪录，也可能一条纪录也没有（不包括第一天）

返回的记录集是一个100天*100个用户的纪录集

下面是我的思路：

1.创建表并插入测试数据：我们要求username从1-100
create table [dbo].[table2] (
[username] [varchar] (50) not null , --用户名
[outdate] [datetime] not null , --日期
[cash] [float] not null --余额
) on [primary

declare @i int
set @i=1
while @i<=100
begin
insert table2 values(convert(varchar(50),@i),'2001-10-1',100)
insert table2 values(convert(varchar(50),@i),'2001-11-1',50)
set @[email protected]+1
end
insert table2 values(convert(varchar(50),@i),'2001-10-1',90)

select * from table2 order by outdate,convert(int,username)

2.组合查询语句：
a.我们必须返回一个从第一天开始到100天的纪录集：
如：2001-10-1（这个日期是任意的）到 2002-1-8
由于第一天是任意一天，所以我们需要下面的sql语句：
select top 100 dateadd(d,convert(int,username)-1,min(outdate)) as outdate
from table2
group by username
order by convert(int,username)
这里的奥妙在于：
convert(int,username)-1（记得我们指定用户名从1-100 :-))
group by username,min(outdate):第一天就可能每个用户有多个纪录。
返回的结果：
outdate
------------------------------------------------------
2001-10-01 00:00:00.000
.........
2002-01-08 00:00:00.000

b.返回一个所有用户名的纪录集：
select distinct username from table2
返回结果：
username
--------------------------------------------------
1
10
100
......
99

c.返回一个100天记录集和100个用户记录集的笛卡尔集合：
select * from
(
select top 100 dateadd(d,convert(int,username)-1,min(outdate)) as outdate
from table2
group by username
order by convert(int,username)
) as a
cross join
(
select distinct username from table2
) as b
order by outdate,convert(int,username)
返回结果100*100条纪录：
outdate username
2001-10-01 00:00:00.000 1
......
2002-01-08 00:00:00.000 100

d.返回当前所有用户在数据库的有的纪录：
select outdate,username,min(cash) as cash from table2
group by outdate,username

order by outdate,convert(int,username)
返回纪录：
outdate username cash
2001-10-01 00:00:00.000 1 90
......
2002-01-08 00:00:00.000 100 50

e.将c中返回的笛卡尔集和d中返回的纪录做left join:
select c.outdate,c.username,
d.cash
from
(
select * from
(
select top 100 dateadd(d,convert(int,username)-1,min(outdate)) as outdate
from table2
group by username
order by convert(int,username)
) as a
cross join
(
select distinct username from table2
) as b
) as c
left join
(
select outdate,username,min(cash) as cash from table2
group by outdate,username
) as d
on(c.username=d.username and datediff(d,c.outdate,d.outdate)=0)

order by c.outdate,convert(int,c.username)
注意：用户在当天如果没有纪录，cash字段返回null，否则cash返回每个用户当天的余额
outdate username cash
2001-10-01 00:00:00.000 1 90
2001-10-01 00:00:00.000 2 100
......
2001-10-02 00:00:00.000 1 90
2001-10-02 00:00:00.000 2 null <--注意这里
......

2002-01-08 00:00:00.000 100 50

f.好了，现在我们最后要做的就是，如果cash为null，我们要返回小于当前纪录日期的第一个用户余额(由于我们使用order by cash,所以返回top 1纪录即可，使用min应该也可以)，这个余额即为当前的余额：
case isnull(d.cash,0)
when 0 then
(
select top 1 cash from table2 where table2.username=c.username
and datediff(d,c.outdate,table2.outdate)<0
order by table2.cash
)
else d.cash
end as cash

g.最后组合的完整语句就是
select c.outdate,c.username,
case isnull(d.cash,0)
when 0 then
(
select top 1 cash from table2 where table2.username=c.username
and datediff(d,c.outdate,table2.outdate)<0
order by table2.cash
)
else d.cash
end as cash
from
(
select * from
(
select top 100 dateadd(d,convert(int,username)-1,min(outdate)) as outdate
from table2
group by username
order by convert(int,username)
) as a
cross join
(
select distinct username from table2
) as b
) as c
left join
(
select outdate,username,min(cash) as cash from table2
group by outdate,username
) as d
on(c.username=d.username and datediff(d,c.outdate,d.outdate)=0)

order by c.outdate,convert(int,c.username)

返回结果：
outdate                            username cash
2001-10-01 00:00:00.000    1         90
2001-10-01 00:00:00.000    2        100
......
2002-01-08 00:00:00.000    100                50

大家看看还有没什么bug，如果你发现bug或者你有更好的方法，你可能发邮件给我：[email protected] ^-^

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