oracle复杂查询练习题
2012-10-17 09:54:16

oracle复杂查询练习题

1.删除重复记录(当表中无主键时)

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Sql代码

create table TESTTB(

bm varchar(4),

mc varchar2(20)

)

insert into TESTTB values(1,'aaaa');

insert into TESTTB values(1,'aaaa');

insert into TESTTB values(2,'bbbb');

insert into TESTTB values(2,'bbbb');

/*方案一*/

delete from TESTTB where rowid not in

(select max(rowid) from TESTTB group by TESTTB.BM,TESTTB.MC)

/*方案二*/

delete from TESTTB a where a.rowid!= (

select max(rowid) from TESTTB b where a.bm=b.bm and a.mc=b.mc

)

2.bookEnrol是用来登记的，不管你是借还是还，都要添加一条记录。

ID为3的java书，由于以归还，所以不要查出来。要求查询结果应为：(被借出的书和被借出的日期)

Sql代码

create table book(

id int ,

name varchar2(30),

PRIMARY KEY (id)

)

insert into book values(1,'English');

insert into book values(2,'Math');

insert into book values(3,'JAVA');

create table bookEnrol(

id int,

bookId int,

dependDate date,

state int,

FOREIGN KEY (bookId) REFERENCES book(id) ON DELETE CASCADE

)

insert into bookEnrol values(1,1,to_date('2009-01-02','yyyy-mm-dd'),1);

insert into bookEnrol values(2,1,to_date('2009-01-12','yyyy-mm-dd'),2);

insert into bookEnrol values(3,2,to_date('2009-01-14','yyyy-mm-dd'),1);

insert into bookEnrol values(4,1,to_date('2009-01-17','yyyy-mm-dd'),1);

insert into bookEnrol values(5,2,to_date('2009-02-14','yyyy-mm-dd'),2);

insert into bookEnrol values(6,2,to_date('2009-02-15','yyyy-mm-dd'),1);

insert into bookEnrol values(7,3,to_date('2009-02-18','yyyy-mm-dd'),1);

insert into bookEnrol values(8,3,to_date('2009-02-19','yyyy-mm-dd'),2);

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/*方案一*/

select a.id,a.name,b.dependdate from book a,bookenrol b where

a.id=b.bookid

and

b.dependdate in(select max(dependdate) from bookenrol group by bookid )

and b.state=1

/*方案二*/

select k.id,k.name,a.dependdate

from bookenrol a, BOOK k

where a.id in (select max(b.id) from bookenrol b group by b.bookid)

and a.state = 1

and a.bookid = k.id;

3.查询每年销量最多的产品的相关信息

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Sql代码

create table t2 (

year_ varchar2(4),

product varchar2(4),

sale    number

)

insert into t2 values('2005','a',700);

insert into t2 values('2005','b',550);

insert into t2 values('2005','c',600);

insert into t2 values('2006','a',340);

insert into t2 values('2006','b',500);

insert into t2 values('2007','a',220);

insert into t2 values('2007','b',350);

insert into t2 values('2007','c',350);

/**方案一*/

select a.year_,a.sale,a.product from t2 a inner join(

select max(sale) as sl from t2 group by year_) b

on a.sale=b.sl  order by a.year_

/*方案二*/

select sa.year_, sa.product, sa.sale

from t2 sa,

(select t.year_ pye, max(t.sale) maxcout

from t2 t

group by t.year_) tmp

where sa.year_ = tmp.pye

and sa.sale = tmp.maxcout

4.排序问题，如果用总积分做降序排序..因为总积分是字符型，所以排出来是这样子(9,8,7,6,5...)，要求按照总积分的数字大小排序。

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Sql代码

create table t4(

)

insert into t4 values('WhatIsJava','1','99');

insert into t4 values('水王','76','981');

insert into t4 values('新浪网','65','96');

insert into t4 values('牛人','22','9');

insert into t4 values('中国队','64','89');

insert into t4 values('信息','66','66');

insert into t4 values('太阳','53','66');

insert into t4 values('中成药','11','33');

insert into t4 values('西洋参','257','26');

insert into t4 values('大拿','33','23');

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/*方案一*/

select * from t4 order by cast(总积分 as int) desc

/*方案二*/

select * from t4 order by to_number(总积分) desc;

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5.得出所有人（不区分人员）每个月及上月和下月的总收入

Sql代码

create table t5 (  tmonth int,

tname varchar2(10),

income number

)

insert into t5 values('08','a',1000);

insert into t5 values('09','a',2000);

insert into t5 values('10','a',3000);

/*方案一*/

select o.tmonth,sum(o.income) as cur,(select sum(t.income) from t5 t where t.tmonth=(o.tmonth+1) group by t.tmonth) as next,

(select sum(t.income) from t5 t where t.tmonth=(o.tmonth-1) group by t.tmonth) as last

from t5 o where o.tmonth=2 group by o.tmonth

/*方案二*/

select tmonth as 月份 ,tname as 姓名,sum(income) as 当月工资,

(select sum(income)

from t5

where tmonth = to_number(substr(to_char(sysdate,'yyyy-mm-dd'),7,1))-1) AS 上月工资 ,

(select sum(income)

from t5

where tmonth = to_number(substr(to_char(sysdate,'yyyy-mm-dd'),7,1))+1) AS 下月工资

from t5 where tmonth=substr(to_char(sysdate,'yyyy-mm-dd'),7,1)

group by tmonth,tname

6.根据现有的学生表，课程表，选课关系表，查询一。没有修过李明老师的课的学生，查询二，既学过a课程，又学过b课程的学生姓名

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Sql代码

S表    [SNO,SNAME]--学生表

C表    [CNO,CNAME,CTEATHER] --课程表

select sname from s where not exists

(select*from sc,c where sc.cno=c.cno and c.cteather='李明' and sc.sno=s.sno)

SELECT S.SNO,S.SNAME

FROM S,(

SELECT SC.SNO

FROM SC,C

WHERE SC.CNO=C.CNO

AND C.CNAME IN('a','b')

GROUP BY SNO

)SC WHERE S.SNO=SC.SNO

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FROM S,SC,(

SELECT SNO

FROM SC

GROUP BY SNO

HAVING COUNT(DISTINCT CNO)>=2

)A WHERE S.SNO=A.SNO AND SC.SNO=A.SNO

GROUP BY S.SNO,S.SNAME