作者:赵磊
博客:http://elf8848.iteye.com
------------------------------------ 练习题一删除重复记录---------------------------------------
前提说明:
distinct 只能在查询时过滤,不能完成本题要的删除功能。 例如 select distinct * from t6
表的结构:
create table t6( bm char(4), mc varchar2(20) )
表中的数据:
insert into t6 values(1,'aaaa'); insert into t6 values(1,'aaaa'); insert into t6 values(2,'bbbb'); insert into t6 values(2,'bbbb');
要求:
删除重复记录
答案:
方法一,好理解:
delete from t6 where rowid not in ( select max(t6.rowid) from t6 group by t6.bm,t6.mc); --这里用min(rowid)也可以。
方法二,比前面的复杂一点:
delete from t6 a where a.rowid!= ( select max(rowid) from t6 b where a.bm=b.bm and a.mc=b.mc )
-------------------------------------- 练习题二 一个查询的练习------------------------------------
表的结构:
表1:book表,字段有id(主键),name (书名);
表2:bookEnrol表(图书借出归还登记),字段有id,bookId(外键),dependDate(变更时间),state(1.借出 2.归还)。
创建表的DDL:
create table book( id int , name varchar2(30), PRIMARY KEY (id) ) create table bookEnrol( id int, bookId int, dependDate date, state int, FOREIGN KEY (bookId) REFERENCES book(id) ON DELETE CASCADE )
表中的数据:
book表:
id name
1 English
2 Math
3 JAVA
bookEnrol表:
id bookId dependDate state
1 1 2009-01-02 1
2 1 2009-01-12 2
3 2 2009-01-14 1
4 1 2009-01-17 1
5 2 2009-02-14 2
6 2 2009-02-15 1
7 3 2009-02-18 1
8 3 2009-02-19 2
插入数据的SQL语句:
insert into book values(1,'English'); insert into book values(2,'Math'); insert into book values(3,'JAVA'); 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);
要求查询出: 不要使用存储过程
第二个表(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)
Id Name dependDate
1 English 2009-01-17
2 Math 2009-02-15
答案:
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
如果大家有更好的方法,可以发出来交流一下。
下面是群友发出来的答案:
副主任的答案
select a.bid, bo.name, a.adate from (select t.bookid bid, t.state, count(*) acou, max(t.dependdate) adate from bookenrol t group by t.bookid, t.state having t.state = 1) a, (select t.bookid bid, t.state, count(*) bcou, max(t.dependdate) bdate from bookenrol t group by t.bookid, t.state having t.state = 2) b, book bo where a.bid = b.bid(+) and bo.id = a.bid and a.acou <> nvl(b.bcou,0);
副主任小师姝的答案
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;
小米的答案
select a.id,a.name,b.dependdate from book a, bookenrol b,(select max(dependdate) dependdate from bookenrol group by bookid) c where a.id(+) =b.bookid and b.dependdate(+)=c.dependdate and b.state=1;
----------------------------------------- 练习题三 一个查询的练习--------------------------------------
表的结构与数据:
表一:各种产品年销售量统计表 sale
年 产品 销量
2005 a 700
2005 b 550
2005 c 600
2006 a 340
2006 b 500
2007 a 220
2007 b 350
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);
要求查询出:
要求得到的结果应为:
年 产品 销量
2005 a 700
2006 b 500
2007 b 350
即:每年销量最多的产品的相关信息。
答案:
我的: select * from t2 a inner join( select year_,max(sale) as sl from t2 group by year_) b on a.year_=b.year_ and a.sale=b.sl 副主任的: 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
-------------------------------------------- 练习题四排序问题-------------------------------------
表的结构与数据:
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');
要求查询出:
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。
答案:
我的: select * from t4 order by cast(总积分 as int) desc 小米的: select * from t4 order by to_number(总积分) desc;
---------------------------------------- 练习题五查询--------------------------------------
表的结构与数据:
表A字段如下
month name income
月份 人员 收入
8 a 1000
9 a 2000
10 a 3000
create table t5 ( month int, name 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); 这个日期要与你做题的日期 相符,才会有当前月,上一个月,下一个月, 过期请自行修改
要求查询出:
要求用一个SQL语句(注意是一个)得出所有人(不区分人员)每个月及上月和下月的总收入
要求列表输出为:
月份 当月收入 上月收入 下月收入
---------- ---------- ---------- ----------
9 2000 1000 3000
答案:
小程的答案:
select o.month,sum(o.income) as cur,(select sum(t.income) from t5 t where t.month=(o.month+1) group by t.month) as next, (select sum(t.income) from t5 t where t.month=(o.month-1) group by t.month) as last from t5 o where o.month=2 group by o.month
重庆--小彭:
select month as 月份 ,name as 姓名,sum(income) as 当月工资, (select sum(income) from t5 where month = to_number(substr(to_char(sysdate,'yyyy-mm-dd'),7,1))-1) AS 上月工资 , (select sum(income) from t5 where month = to_number(substr(to_char(sysdate,'yyyy-mm-dd'),7,1))+1) AS 下月工资 from t5 where month=substr(to_char(sysdate,'yyyy-mm-dd'),7,1) group by month,name
drop table t5 create table t5 ( month date, name varchar2(10), income number ) insert into t5 values(to_date('2010-08-01','yyyy-MM-dd'),'a',1000); insert into t5 values(to_date('2010-09-01','yyyy-MM-dd'),'a',2000); insert into t5 values(to_date('2010-10-01','yyyy-MM-dd'),'a',3000); select sum(to_number(substr(to_char(sysdate,'yyyy-MM-dd'),6,2)))/count(*), sum(decode(month, to_date(to_char(add_months(trunc(sysdate),-1),'yyyy-MM'),'yyyy-MM'), income, 0)) 上月, sum(decode(month, to_date(to_char(add_months(trunc(sysdate),0),'yyyy-MM'),'yyyy-MM'), income, 0)) 当月, sum(decode(month, to_date(to_char(add_months(trunc(sysdate),1),'yyyy-MM'),'yyyy-MM'), income, 0)) 下月 from t5 group by name;
----------------------------------------------- 练习题其它----------------------------------------
rowid 和 rownum 的区别
rowid 用于定位数据表中某条数据的位置,是唯一的、也不会改变
rownum 表示查询某条记录在整个结果集中的位置,
同一条记录查询条件不同对应的 rownum 是不同的而 rowid 是不会变的