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SQL查询面试题

2012-11-09 22:43 357 查看
1、统计每年每月的信息

year month amount

1991 1 1.1

1991 2 1.2

1991 3 1.3

1991 4 1.4

1992 1 2.1

1992 2 2.2

1992 3 2.3

1992 4 2.4

查成这样一个结果

year m1 m2 m3 m4

1991 1.1 1.2 1.3 1.4

1992 2.1 2.2 2.3 2.4
准备sql语句:

drop table if exists sales;

create table sales(id int auto_increment primary key,year varchar(10), month varchar(10), amount float(2,1));

insert into sales values

(null,’1991′,’1′,1.1),

(null,’1991′,’2′,1.2),

(null,’1991′,’3′,1.3),

(null,’1991′,’4′,1.4),

(null,’1992′,’1′,2.1),

(null,’1992′,’2′,2.2),

(null,’1992′,’3′,2.3),

(null,’1992′,’4′,2.4);

答案:

select sales.year ,

(select t.amount from sales t where t.month=’1′ and t.year= sales.year) ’1′,

(select t.amount from sales t where t.month=’1′ and t.year= sales.year) ’2′,

(select t.amount from sales t where t.month=’1′ and t.year= sales.year) ’3′,

(select t.amount from sales t where t.month=’1′ and t.year= sales.year) as ’4′

from sales group by year;

2、分组统计胜负

表内容:

2005-05-09 胜

2005-05-09 胜

2005-05-09 负

2005-05-09 负

2005-05-10 胜

2005-05-10 负

2005-05-10 负

如果要生成下列结果, 该如何写sql语句?

胜 负

2005-05-09 2 2

2005-05-10 1 2

------------------------------------------

create table #tmp(rq varchar(10),shengfu nchar(1))

insert into #tmp values('2005-05-09','胜')

insert into #tmp values('2005-05-09','胜')

insert into #tmp values('2005-05-09','负')

insert into #tmp values('2005-05-09','负')

insert into #tmp values('2005-05-10','胜')

insert into #tmp values('2005-05-10','负')

insert into #tmp values('2005-05-10','负')

1)select rq, sum(case when shengfu='胜' then 1 else 0 end)'胜',sum(case when shengfu='负' then 1 else 0 end)'负' from #tmp group by rq

2) select N.rq,N.勝,M.負 from (

select rq,勝=count(*) from #tmp where shengfu='胜'group by rq)N inner join

(select rq,負=count(*) from #tmp where shengfu='负'group by rq)M on N.rq=M.rq

3)select a.col001,a.a1 胜,b.b1 负 from

(select col001,count(col001) a1 from temp1 where col002='胜' group by col001) a,

(select col001,count(col001) b1 from temp1 where col002='负' group by col001) b

where a.col001=b.col001
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