01. tmp表中有如下记录（建表 SQL见emp.sql）
RQ SHENGFU
---------- ----------
2005-05-09 WIN
2005-05-09 WIN
2005-05-09 LOSE
2005-05-09 LOSE
2005-05-10 WIN
2005-05-10 LOSE
2005-05-10 LOSE
要求格式为：
RQ WIN LOSE
---------- ---------- ----------
2005-05-10 1 2
2005-05-09 2 2
答案：select rq, count(case when shengfu='WIN' then 'WIN' else null end) WIN, count(case when shengfu='LOSE' then 'LOSE' else null end) LOSE from tmp group by rq ;
02. 查询当前月有多少天
答案：SQL> select trunc(add_months(sysdate,1),'month') - trunc(sysdate,'month') from dual;
03. pages表有四个字段，id, url,title,body。如图：
ID URL TITLE BODY
--------- ----------- ------------------------- ------
1 http://www.baidu.com 新闻 党报评事业单位发绩效工资 砸铁饭碗再砸金饭碗
2 http://www.sina.com baidu新闻 假唱假演奏最高罚款3000元 10月1日起施行
3 http://www.yahoo.com 搜索结果 www.baidu.com/search/url_submit.html - 网页快
4 http://www.baidu.com 新闻 垃圾焚烧产生致癌物 专家告诫中国勿重蹈日本覆辙
要求格式为：
ID CUNT
--------------------------------------------------------------------------------
3 www.baidu.com/search/url_submit.html - 网页快照 - 类似结果
2 baidu新闻
1 http://www.baidu.com
4 http://www.baidu.com
答案：select id,body cunt from pages where body like '%baidu%' union all select id,title from pages where title like '%baidu%' union all select id,url from pages where url like '%baidu%';
ID URL
--------------------------------------------------------------------------------
1 http://www.baidu.com
4 http://www.baidu.com
2 baidu新闻
3 www.baidu.com/search/url_submit.html - 网页快照 - 类似结果
答案：select id,url from pages where url like '%baidu%' union all select id,title from pages where title like '%baidu%' union all select id,body cunt from pages where body like '%baidu%';
04. 现有 STUDENT(学生), COURSE(课程), SC(成绩)表,完成以下需求(建表语句在 emp.sql
中,综合考察)
a) 查询选修课程为 web的学员学号和姓名
答案：SQL> Select s.name,s.sid from student s, (Select sid from sc Where cid=(select cid from course where name='web' )) s1 where s.sid=s1.sid;
b) 查询课程编号为 2的学员姓名和单位
答案：select * from student s,(select sid from sc where cid=2) s1 where s.sid=s1.sid;
b) 查询不选修4号课程的学员姓名和单位
答案：SQL> select distinct s.* from student s,(Select sid from sc where cid!=4) s1 where s.sid=s1.sid;
d) 查询选修全部课程的学员姓名和单位
答案：SQL> select * from student where sid=(select sid from sc group by sid having count(*)=(select count(distinct cid) from sc));
e) 查询选修课程超过 3门的学员姓名和单位
答案：SQL> select * from student s,(select sid from sc group by sid having count(*)>3) s1 where s.sid=s1.sid;
f) 找出没有选修过 Teacher LI讲授课程的所有学生姓名
答案：select distinct s1.* from student s1 where s1.sid not in ( select s.sid from sc s,(select distinct cid from course where TEACHER='Teacher LI') c where s.cid=c.cid) ;
g) 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
答案：SQL> select s.*,sco from student s,(select distinct sid,avg(SCORE) sco from sc where score<60 group by sid) s1 where s.sid=s1.sid;
h) 列出既学过1号课程，又学过2号课程的所有学生姓名
答案：SQL> select * from student s, (select sid from sc where cid=1) s1 ,(select sid from sc where cid=2) s2 where s.sid=s1.sid and s.sid=s2.sid;
i) 列出1号课成绩比2号课成绩高的所有学生的学号，姓名和 1号课和 2号课的成
绩
答案：select * from student s, (select sid,SCORE from sc where cid=1) s1 ,(select sid,SCORE from sc where cid=2) s2 where s.sid=s1.sid and s.sid=s2.sid and s1.score>s2.score;
05. 现有test表，表中数据如图所示：
a) 连续的编号要求如下格式
BEGIN END
---------- ----------
1 6
8 9
11 14
18 19
答案：Select min(id) begin,max(id) end from test group by id-rownum order by id-rownum
b) 不连续的编号要求如下格式
BEGIN END
---------- ----------
8 9
11 14
18 19
答案：SQL> Select min(id) begin,max(id) end from test group by id-rownum having id-rownum!=0 order by id-rownum;
06.（统计各部门，各职位的人数）
DEPTNO CLERK SALESMAN PRESIDENT MANAGER ANALYST
------ ---------- ---------- ---------- ---------- ----------
30 1 4 0 1 0
20 2 0 0 1 2
10 1 0 1 1 0
答案：select deptno, count(case when job='CLERK' then 'CLERK' else null end) CLERK, count(case when job = 'SALESMAN' then 'SALESMAN' else null end) SALESMAN, count(case when job='PRESIDENT' then 'PRESIDENT' else null end) PRESIDENT, count(case when job='MANAGER' then 'MANAGER' else null end) MANAGER, count(case when job='ANALYST' then 'ANALYST' else null end) ANALYST from emp group by deptno;
07. 根据EMP表数据产生如下格式的报表(统计各职位,各部门的人数)(06题的变体)
Job 10 20 30
------------- ---------- ----------- ---------
0 0 1
CLERK 1 2 1
SALESMAN 0 0 4
PRESIDENT 1 0 0
MANAGER 1 1 1
ANALYST 0 2 0
答案：select job , count(case when deptno='10' then '10' else null end) as "10", count(case when deptno='20' then '20' else null end) as "20", count(case when deptno='30' then '30' else null end) as "30" from emp group by job;
08. 按照如下格式显示 7369号员工的信息
Empno key value
------------ ------------ ---------------------------
7369 comm
7369 deptno 20
7369 ename smith
7369 hiredate 1980-12-17
7369 job cleak
7369 mgr 7902
7369 sal 800
答案：select empno, 'ENAME' as KEY, ename VALUE from emp where empno = 7369
union
select empno, 'JOB', job from emp where empno = 7369
union
select empno, 'HIREDATE', to_char(hiredate,'yyyy-mm-dd') a from emp where empno = 7369
union
select empno, 'MGR', to_char(mgr) from emp where empno = 7369
union
select empno, 'SAL', to_char(sal) from emp where empno = 7369
union
select empno, 'COMM', to_char(comm) from emp where empno = 7369
union
select empno, 'DEPTNO', to_char(deptno) from emp where empno = 7369;
oracle常见的经典查询语句（二）
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转载自ymq267.iteye.com/blog/2247736
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