sql 练习（1）

　　1.建立实验表

  CREATE TABLE STUDENT
 (SNO VARCHAR2(3) NOT NULL,
    SNAME VARCHAR2(40) NOT NULL,
    SSEX VARCHAR2(20) NOT NULL,
    SBIRTHDAY DATE,
    CLASS VARCHAR2(20))
 /
 CREATE TABLE COURSE
 (CNO VARCHAR2(5) NOT NULL,
    CNAME VARCHAR2(10) NOT NULL,
    TNO VARCHAR2(10) NOT NULL)
 /
 CREATE TABLE SCORE
 (SNO VARCHAR2(3) NOT NULL,
    CNO VARCHAR2(5) NOT NULL,
    DEGREE NUMERIC(10, 1) NOT NULL)
 /
 CREATE TABLE TEACHER
 (TNO VARCHAR2(3) NOT NULL,
    TNAME VARCHAR2(20) NOT NULL, TSEX VARCHAR2(20) NOT NULL,
    TBIRTHDAY DATE , PROF VARCHAR2(60),
    DEPART VARCHAR2(10) NOT NULL)
 /
 INSERT INTO STUDENT (SNO,SNAME,SSEX,SBIRTHDAY,CLASS) VALUES (108 ,'tom','man','',95033);
 INSERT INTO STUDENT (SNO,SNAME,SSEX,SBIRTHDAY,CLASS) VALUES (105 ,'jon','man','',95031);
 INSERT INTO STUDENT (SNO,SNAME,SSEX,SBIRTHDAY,CLASS) VALUES (107 ,'lily','woman','',95033);
 INSERT INTO STUDENT (SNO,SNAME,SSEX,SBIRTHDAY,CLASS) VALUES (101 ,'mac','man','',95033);
 INSERT INTO STUDENT (SNO,SNAME,SSEX,SBIRTHDAY,CLASS) VALUES (109 ,'mary','woman','',95031);
 INSERT INTO STUDENT (SNO,SNAME,SSEX,SBIRTHDAY,CLASS) VALUES (103 ,'fuck','man','',95031);

 INSERT INTO COURSE(CNO,CNAME,TNO)VALUES ('3-105' ,'computer science',825);
 INSERT INTO COURSE(CNO,CNAME,TNO)VALUES ('3-245' ,'operate system' ,804);
 INSERT INTO COURSE(CNO,CNAME,TNO)VALUES ('6-166' ,'digital circal' ,856);
 INSERT INTO COURSE(CNO,CNAME,TNO)VALUES ('9-888' ,'maths' ,100);

 INSERT INTO SCORE(SNO,CNO,DEGREE)VALUES (103,'3-245',86);
 INSERT INTO SCORE(SNO,CNO,DEGREE)VALUES (105,'3-245',75);
 INSERT INTO SCORE(SNO,CNO,DEGREE)VALUES (109,'3-245',68);
 INSERT INTO SCORE(SNO,CNO,DEGREE)VALUES (103,'3-105',92);
 INSERT INTO SCORE(SNO,CNO,DEGREE)VALUES (105,'3-105',88);
 INSERT INTO SCORE(SNO,CNO,DEGREE)VALUES (109,'3-105',76);
 INSERT INTO SCORE(SNO,CNO,DEGREE)VALUES (101,'3-105',64);
 INSERT INTO SCORE(SNO,CNO,DEGREE)VALUES (107,'3-105',91);
 INSERT INTO SCORE(SNO,CNO,DEGREE)VALUES (108,'3-105',78);
 INSERT INTO SCORE(SNO,CNO,DEGREE)VALUES (101,'6-166',85);
 INSERT INTO SCORE(SNO,CNO,DEGREE)VALUES (107,'6-106',79);
 INSERT INTO SCORE(SNO,CNO,DEGREE)VALUES (108,'6-166',81);

 INSERT INTO TEACHER(TNO,TNAME,TSEX,TBIRTHDAY,PROF,DEPART)
 VALUES (804,'Mr Li','man','','js','computer');
 INSERT INTO TEACHER(TNO,TNAME,TSEX,TBIRTHDAY,PROF,DEPART)
 VALUES (856,'Mrs Zhang','man','','te','elecitric');
 INSERT INTO TEACHER(TNO,TNAME,TSEX,TBIRTHDAY,PROF,DEPART)
 VALUES (825,'Mrs Wang','woman','','ta','computer');
 INSERT INTO TEACHER(TNO,TNAME,TSEX,TBIRTHDAY,PROF,DEPART)

　　2.查询Score表中的最高分的学生学号和课程号。
select sno,cno from score where degree=(select max(degree) from score);

　　3.查询Score表中至少有5名学生选修的并以3开头的课程的平均分数。
统计用分组:
select avg(degree) from score group by cno having count(cno)>5 and con like '3-%';

　　4.create table grade(low   number(3,0),upp   number(3),rank   varchar(10));

SQL> insert into grade values(90,100,'A');

1 row created.

SQL> insert into grade values(80,89,'B');

1 row created.

SQL> insert into grade values(70,79,'C');

1 row created.

SQL> insert into grade values(60,69,'D');

1 row created.

SQL> insert into grade values(0,59,'E');

1 row created.

现查询所有同学的Sno、Cno和rank列。
计算使用子查询传递参数；
select Sno,Cno,rank from score,grade where degree between low and upp;
select sno,cno,(select rank from grade where sc.degree>low and sc.degree<upp) rank from score sc;

　　5.查询score中选学两门以上课程的同学中分数为非最高分成绩的记录。
把限制拆开来看；
select * from score where degree not in (select max(degree) from score group by cno) and cno not in (select cno from score group by cno having count(*)<=2);

　　6.查询选修某课程的同学人数多于5人的教师姓名。

多表分组（函数）转化为子查询分组；
select tname from teacher where tno in (select tno from score sc,course c where sc.cno=c.cno group by tno having count(*)>5);

　　7.查询成绩比该课程平均成绩低的同学的成绩表。
select sname ,cname ,degree from student s,score sc,course c where sc.sno=s.sno and c.cno=sc.cno and degree<(select avg(degree) from score scc where scc.cno=sc.cno);

　　8.查询每个班有哪些老师
等价挂载点问题；
班级转化为班里面的sno
老师转化为cno
然后加distinct即为一个班对应一个sno，一个老师对应一个cno

　　9.查询“c001”课程比“c002”课程成绩高的所有学生的学号；
select * from score a,score b where a.sno=b.sno and a.cno='3-105' and b.cno='3-245' and a.degree>b.degree;
select * from score a where a.sno in (select b.sno from score b where a.cno='3-105' and a.sno=b.sno and b.cno='3-245' and a.degree>b.degree);

　　10.查询没有学全所有课的同学的学号、姓名；
集合做差：

SQL> select * from student where sno in (select sno from (select stu.sno,c.cno from student stu cross join course c minus select sno,cno from score));

　　11.查询和“s001”号的同学学习的课程完全相同的其他同学学号和姓名；
集合操作：构造伪满集；

　　12.按各科平均成绩从低到高和及格率的百分数从高到低顺序
select avg(degree) ,(sum(case when degree>60 then 1 else 0 end)/count(*)) jige from score group by cno;

　　13.查询各科成绩前三名的记录:(不考虑成绩并列情况)
按内部分组编号：
select sno,cno,degree,row_number() over (partition by cno order by degree desc) rn from score;

　　14.查询全部学生都选修的课程的课程号和课程名
伪满集操作

　　15.列出至少有一个雇员的所有部门
内连接的定义：部门表要在雇员表中出现：俩个表中有相互联系的数据

　　16.列出按年薪排序的所有雇员的年薪
select (sal+nvl(comm,0))*12 as avn from emp order by avn

　　17.列出薪金水平处于第四位的雇员
Select * from (Select ename,sal, rank() over (order by sal desc) as grade from emp) where grade=4

　　18.找出不收取佣金或收取的佣金低于100的雇员
select * from emp where nvl(comm,0)<100;

　　19.找出各月最后一天受雇的所有雇员
select * from emp where hiredate= last_day(hiredate);

　　20.找出早于25年之前受雇的雇员
select * from emp where months_between(sysdate,hiredate)/12>25;
select * from emp where hiredate<add_months(sysdate,-12*25);

　　21显示只有首字母大写的所有雇员的姓名
select ename from emp where ename=initcap(ename);

　　22.显示不带有'R'的雇员姓名
Select ename from emp where ename not like ‘%R%’;
Select ename from emp where instr(ename,’R’)=0;

　　23.以年、月和日显示所有雇员的服务年限
Select months_between(sysdate,hiredate)/12 as “年”, months_between(sysdate,hiredate) as “月”, sysdate-hiredate as “日” from emp

　　24.显示每个员工每天是否有迟到和早退；
8:00--12:00 为迟到, 12:00--18:00为早退

 打卡表  card
 SQL> create table card(
     cid number(20),
     ctime date,
     cuser number(20)); 

 人员表  person
 create table person(
  pid number(20),
  name varchar2(10)
 )
  --插入人员表的数据
 insert into person values(1,'a');
 insert into person values(2,'b'); 

 --插入打卡的数据
 insert into card values(1,to_date('','yyyymmddhh24miss'),1);
 insert into card values(2,to_date('','yyyymmddhh24miss'),1);
 insert into card values(3,to_date('','yyyymmddhh24miss'),2);
 insert into card values(4,to_date('','yyyymmddhh24miss'),2); 

 insert into card values(5,to_date('','yyyymmddhh24miss'),1);
 insert into card values(6,to_date('','yyyymmddhh24miss'),1);
 insert into card values(7,to_date('','yyyymmddhh24miss'),2);
 insert into card values(8,to_date('','yyyymmddhh24miss'),2); 

 --分析：  先分组统计出每个人，每天的上班时间和下班时间  即（id,day,mindate,maxdate)
 select  p.pid as id,
  to_char(c.ctime,'yyyymmdd') as day,
  to_char(min(c.ctime),'hh24mi') as mindate,
  to_char(max(c.ctime),'hh24mi') as maxdate
  from card c,person p where c.cuser = p.pid group by p.pid,to_char(c.ctime,'yyyymmdd');
  --把上面的分析做成一个视图，判断上班时间是否为迟到  和  下班时间是否为早退
 -- 如 果 判 断 前 10 天 的 打 卡 记 录 ， 就   改成
 to_char(c.ctime,'yyyymmdd')<=to_char(sysdate-10,'yyyymmdd') 

 select  p.name as person_name,
  e1.day as work_day,
  e1.mindate as AM,
  e1.maxdate as PM,
  --判断迟到
  case
   when e1.mindate between '' and '' then 'yes'
   else 'no'
  end as later,
  --判断早退
  case
   when e1.maxdate between '' and '' then 'yes'
   else 'no'
  end as leave_early
  from
   --员工表
   person p,
   --上面那张视图表
   (select
    p.pid as id,
    to_char(c.ctime,'yyyymmdd') as day,
    to_char(min(c.ctime),'hh24mi') as mindate,
    to_char(max(c.ctime),'hh24mi') as maxdate
       from card c,person p
    where
     c.cuser = p.pid and
     to_char(c.ctime,'yyyymmdd')<=to_char(sysdate-1,'yyyymmdd')
    group by p.pid,to_char(c.ctime,'yyyymmdd')
   ) e1
  where p.pid = e1.id; 

　　25.删除一张表重复记录（ID 是自增唯一，重复记录：其他字段都是一样）
非常经典的一道面试题（可能存在很多数据，要求性能比较高）

1 louis 20
2 louis 20
3 jimmy 30
4 louis 20
------------------------------------------------------------------
delete from aa where id not in(select min(id) from aa group by name,age);

select a1.id
from a a1,
a a2
where a1.id>a2.id and a1.name=a2.name and a1.age=a2.age and a1.sex=a2.sex