oracle复杂查询练习题

1.删除重复记录(当表中无主键时)

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。

请写一个SQL语句，获取到现在状态为已借出的所有图书的相关信息，

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)

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); 

/*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

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。

create table t4(  
姓名   varchar2(20),  
月积分 varchar2(20),  
总积分 char(3)  
)  
  
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');  
 
/*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;

5.得出所有人（不区分人员）每个月及上月和下月的总收入

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

S表    [SNO,SNAME]--学生表
C表    [CNO,CNAME,CTEATHER] --课程表
SC表  [SNO，CNO，SCGRADE] --选课关系表

查询一：没有修过李明老师的课的学生的姓名
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)

查询二：既学过a课程，又学过b课程的学生姓名
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 

查询三： 列出有二门以上（含两门）不及格课程的学生姓名及其平均成绩 
SELECT S.SNO,S.SNAME,AVG(SC.SCGRADE)
FROM S,SC,( 
     SELECT SNO 
     FROM SC 
     WHERE SCGRADE <60 
     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

oracle复杂查询练习题

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