经典SQL数据库面试题以及答案—Oracle版本-SQL全部在plsql开发编写-欢迎提问

Student(Sno,Sname,Sage,Ssex) 学生表       S1:学号;Sname:学生姓名;Sage:学生年龄;Ssex:学生性别
Course(Cno,Cname,T1) 课程表               C1,课程编号;Cname:课程名字;T1:教师编号
SC(Sno,Cno,score) 成绩表                   S1:学号;C1,课程编号;score:成绩
Teacher(Tno,Tname) 教师表                 T1:教师编号; Tname:教师名字
--建表
create table student(
sno int primary key,
sname varchar2(10) not null,
sage int  not null,
sses char(1) ,
constraint check_age check((sage)>0 and (sage)<120)
);

create table course(
cno int  primary key,
cname varchar2(10) not null,
tno int not null,
constraint  fk_course_teacher foreign key(tno) references  teacher(tno)
);

create table teacher(
tno int primary key,
tname varchar2(10) not null
);

create table sc (
sno int ,
cno int ,
score number,
constraint fk_cno_sc foreign key (cno) references course(cno),
constraint fk_cno_student foreign key (sno) references student(sno)
);
--修改表
alter table student rename column sses to ssex;
alter table student drop column ssex;
alter table student add ( ssex varchar2(2)) ;
--插数据
insert into student values('001','zhang三',22,'男');
insert into student values('002','wangwu2',22,'男');
insert into student values('003','史蒂夫',23,'男');
insert into student values('004','艾德',20,'女');
insert into student values('005','猪八戒',32,'男');
insert into student values(6,'合家福',12,'女');
select * from student;

insert into teacher values('1','徐大大');
insert into teacher values('2','王富人');
insert into teacher values('3','大大');
insert into teacher values('5','李大大');
insert into teacher values('6','陈大大');
insert into teacher values('89','刘在石');
insert into teacher values('9','何达大');
insert into teacher values('10','彭建军');
select * from teacher;

insert into course values('1','Oracle数据','1');
insert into course values('2','计算机原理','1');
insert into course values('3','数据结构','3');
insert into course values('4','算法概论','3');
insert into course values('5','数据库原理','3');
insert into course values('6','软件工程','6');
insert into course values('7','UML','5');
insert into course values('8','Java语言','89');
insert into course values('9','嵌入式系统','10');
select * from course order by cno desc;
delete from sc s where rowid = (select  rowid from sc where sno =3 and rownum =1 and cno =8);
delete from sc s where rowid !=(select max(rowid) from sc b where s.cno=b.cno and b.cno =2)
delete from sc s where cno =2;
select rowid from sc where cno = 8;

select cno,count(cno) from sc group by cno having count(cno) !=5 ;/2 4 3

insert into sc(sno,cno,score) values(6,1,48);
insert into sc(sno,cno,score) values(6,2,55);
insert into sc(sno,cno,score) values(6,3,46);
insert into sc(sno,cno,score) values(6,4,35);
insert into sc(sno,cno,score) values(6,5,52);
select * from sc where sno =6;
select count(cno),cno from sc group by cno;
select * from sc order by cno;
select * from sc where cno =4 or cno =3 order by cno ,sno;
select * from sc order by cno asc,sno asc,score desc;

select s.sno,s.Sname,count(sc.sno),sum(sc.score) from student s
left join sc on sc.sno = s.sno group by sc.sno ;
--Question and Exercise
1、查询“001”课程比“002”课程成绩高的所有学生的学号;
select a.sno from 
       (select sno,score from sc where cno =1) a,
       (select sno,score from sc where cno =2) t 
       where a.score >t.score and a.sno =t.sno;
//sno=1 2 3 4 5
select * from sc where sno =5
2、查询平均成绩大于60分的同学的学号和平均成绩;
select sno,round(avg(score),1) from sc group by sno having avg(score)> 60;
//round(数值,小数位数) Math.round()四舍五入取整原理:先+0.5 然后再向下取整(floor ),ceil
//decimal()
3、查询所有同学的学号、姓名、选课数、总成绩;
select s.sno,s.sname,count(c.rowid),sum(c.score) from student s ,sc c group by c.sno; 错误!!! 连接查询 和 分组
select s.sno,s.sname,count(c.cno), sum(c.score) from student s left join sc c on  s.sno = c.sno group by s.sno ,s.sname order by s.sno;
select s.sno,s.sname,s.ssex,count(c.cno),sum(c.score) from student s,sc c where s.sno = c.sno(+) group by s.sno,s.sname,s.ssex;
/*但是现在一旦分组之后,实际上对于语法上就会出现了新的限制,对于分组有以下要求:
1分组函数可以在没有分组的时候单独用使用,可是却不能出现其他的查询字段;
分组函数单独使用:
  SELECT COUNT(empno) FROM emp;
  错误的使用,出现了其他字段:
  SELECT empno,COUNT(empno) FROM emp;
2如果现在要进行分组的话,则SELECT子句之后,只能出现分组的字段和统计函数,其他的字段不能出现:
  正确做法:
  SELECT job,COUNT(empno),AVG(sal)
  FROM emp
  GROUP BY job;
  错误的做法:
  SELECT deptno,job,COUNT(empno),AVG(sal)
  FROM emp
  GROUP BY job;
*/
4、查询姓“李”的老师的个数;
select count(s.tno),s.tname from teacher s where s.tname like '李%' group by tname;

5、查询没学过“叶平”老师课的同学的学号、姓名;
select * from teacher;
insert into teacher values(66,'叶平');
insert into course values(0,'C语言基础',66);
insert into sc values(1,0,98)
嵌套
select t.sno,t.sname from student t where not exists  (
select sno from sc  where t.sno =sc.sno and cno in (
select cno from course where tno in(
select tno from teacher where tname='叶平'
)));
内连接
select t.sno,t.sname from student t where t.sno not in(
select sno from sc s,course c,teacher b where 
s.cno = c.cno and
c.tno = b.tno and
b.tname = '叶平'
);

6、查询学过“001”并且也学过编号“002”课程的同学的学号、姓名;
//select * from sc 
//select s.sno 学号,sname 姓名 from student  s,sc c where s.sno=c.sno(+) and c.cno =1 and c.cno =2;
//select s.sno ,s.sname from student s left join sc c on s.sno = c.sno where c.cno between 1 and 2;
select distinct s.sno , s.sname from student s ,sc c where s.sno = c.sno and c.cno between 1 and 2  group by s.sno,s.sname having count(s.sno)=2;
//select s.sno , s.sname from student where s.sno in ()

7、查询学过“叶平”老师所教的所有课的同学的学号、姓名;
select s.sno,s.sname from student s,sc c,course co ,teacher t
where
s.sno =c.sno and
c.cno =co.cno and
co.tno = t.tno and
t.tname ='叶平';

8、查询课程编号“002”的成绩比课程编号“001”课程低的所有同学的学号、姓名;
select sno,sname from student s where s.sno in(
select a.sno from (select score,sno from sc where cno=1) a
left join (select score,sno from sc where cno=2) b
on a.sno = b.sno where a.score>b.score 
);
update sc set score =99 where cno =2 and sno=1;
select * from sc where cno between 1 and 2 ;
9、查询所有课程成绩小于60分的同学的学号、姓名;--没有一门成绩>60
采用 any some all 类的比较量词
select distinct s.sno,sname from student s left join sc b on s.sno= b.sno where 60> all(select score from sc z where z.sno=b.sno);

select s.sno,s.sname from student s where s.sno not in
(select sno from sc b where b.score >60 and s.sno=b.sno);

10、查询没有学全所有课的同学的学号、姓名;
select * from sc order by sno,cno,score;
select s.sno,s.sname from student s where s.sno in (
select sno from (select sno,count(rowid) from sc group by sno
 having count(rowid)<10 order by sc.sno asc)
);
select s.sno,s.sname from student s,sc b where
s.sno = b.sno group by s.sno,s.sname having count(cno) < (select count(cno) from course);

11、查询至少有一门课与学号为“1001”的同学所学相同的同学的学号和姓名;
select s.sno ,s.sname from student s where s.sno in 
(
select sno from sc b where  exists (select cno from sc c where sno =6 and b.cno =cno)
)
select distinct s.sno ,s.sname from student s,sc b where
 s.sno =b.sno and b.cno in (select cno from sc t where  t.sno=6);
测试 一个没有跟sno=1相同学好的同学
select * from sc where sno =6
insert into student values(7,'刘备',23,'男')
insert into sc values(7,9,'89')

12、查询至少学过学号为“001”同学所有一门课的其他同学学号和姓名;
select distinct s.sno,s.sname from student s,sc b where s.sno = b.sno and b.cno in (select cno from sc where sno = 6);

13、把“SC”表中“叶平”老师教的课的成绩都更改为此课程的平均成绩;
--oracle产生随机数
select round(dbms_random.value(0,100),1) from dual;
dbms_random.string (opt char, len NUMBER);
--opt:表示返回值可能包含的东西
----'u','U'  :  任何大写字母
----'l','L'  :  任何小写字母
----'a','A'  :  大小写混合字母
----'x','X'  :  大写字母和数字混合
----'p','P'  :  任何可打印字符
----其他参数  :  任何大写字母
--len:表示要返回的字符串长度
select dbms_random.string('x', 19) from dual


begin
for cr in (select rowid from sc )
  loop  --循环
    update sc b set score =round(dbms_random.value(0,100),1)    --更新语句(根据查询出来的结果集合)
    where rowid = b.rowid;
  end loop;         --结束循环
end;
commit;
/*sql循环更新模板
BEGIN
FOR L_RECORD IN (select id from customer_config where salt is null) 
LOOP
        UPDATE customer_config
           SET passwd=dbms_random.string('x',8)
         WHERE id=L_RECORD.id;
 END LOOP;
 commit;
END;*/
update sc so set score = (
select distinct round(avg(score),1) from sc b where b.cno = (
select distinct s.cno from sc s,course c,teacher t where s.cno = c.cno and c.tno = t.tno and t.tname = '叶平'
)) where  so.cno = 
( select distinct s.cno from sc s,course c,teacher t where s.cno = c.cno and c.tno = t.tno and t.tname = '叶平')

update sc s1 set score=(
  select avg(sc2.score) from  sc s2 where s1.sno = s2.sno
) from sc sc2,course c,teacher t where sc2.cno =c.cno and t.tno = c.tno and t.tname ='叶平';

insert into sc values(4,0,round(dbms_random.value(0,100),1));
select * from sc b where b.cno = 0;
14、查询和“1002”号的同学学习的课程完全相同的其他同学学号和姓名;
select * from sc order by sno,cno,score;
select distinct s.sno,s.sname from student s,sc b where s.sno = b.sno and 
b.cno in(select cno from sc where sno = 1)


15、删除学习“叶平”老师课的SC表记录;
select * from  sc b,course c,teacher t where b.cno = c.cno and c.tno = t.tno and t.tname ='叶平';
select * from sc where cno = 2;
delete from sc b,course c,teacher t where b.cno = c.cno and c.tno =t.tno and t.tname = '叶平';
select * from course c,teacher t where c.tno = t.tno and t.tname = '叶平';

delete from sc b where b.sno in (select bb.sno from sc bb,course c,teacher t 
where b.cno = bb.cno and bb.cno= c.cno and c.tno = t.tno and t.tname ='叶平')

16、向SC表中插入一些记录,这些记录要求符合以下条件:没有上过编号“003”课程的同学学号、2、
    号课的平均成绩;
    begin
      for sn in(select distinct sno from sc b where b.sno not in (select sno from sc bb where bb.cno=3) )
        loop
          insert into sc values(sn.sno,2,(select round(avg(a.score),1) from sc a group by a.cno having a.cno = 2));
        end loop;
      end;
    
    insert into sc select s.sno,2,(select round(avg(score),1) from sc b where b.cno=2)
    from student s where s.sno not in (select sno from sc where cno =3)
    
    select * from sc where sno =7;
17、按平均成绩从高到低显示所有学生的“数据库”、“企业管理”、
“英语”三门的课程成绩,按如下形式显示: 学生ID,数据库,企业管理,英语,有效课程数,有效平均分
select sno,b.cno,(select cname from course c where cno =b.cno),count(*),avg(score)
from sc b
group by sno,cno

select s1.sno 学生ID,
(select b.score from sc b where b.cno =1 and  b.sno =s1.sno) as 数据库,
(select b.score from sc b where b.cno =8 and  b.sno = s1.sno) Java,
(select b.score from sc b where b.cno =3 and  b.sno = s1.sno) 数据结构,
count(*) 有效课程数,
round(avg(score),1)有效平均分
from sc s1
group by s1.sno
order by round(avg(score),1)

18、查询各科成绩最高和最低的分:以如下形式显示:课程ID,最高分,最低分
select   b.cno 课程ID,
        (select cname from course where b.cno = cno) 课程,
        max(score) 最高分,
        min(score) 最低分,
        round(avg(score),1) 平均分
from sc b group by b.cno


19、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 b.cno 课程编号,
       (select cname from course where cno =b.cno) 课程名,
       round(avg(b.score),1) 平均分,
       concat(round(sum(case when b.score>60 then 1 else 0 end)/count(*)*100,1),'%') 及格率
       from sc b group by b.cno order by avg(b.score);
--验证 cno 的及格率       
       select sum(score) ,round(
       sum(
       case 
         when b.score>=60
           then 1
           else 0
       end
       )/count(*),2) 及格率,count(*) from sc b where cno =2;
       select * from sc where cno = 2 3个及格的,共5个,及格率60%

20、查询如下课程平均成绩和及格率的百分数(用"1行"显示): 企业管理(001),马克思(002),OO&UML (003),数据库(004)
select b.cno 课程编号,
       (select cname from course where cno =b.cno) "课程名",
       round(avg(b.score),1) 平均分,
       concat(round(sum(case when b.score>60 then 1 else 0 end)/count(*)*100,1),'%') 及格率
       from sc b group by b.cno having b.cno in(1,2,3,4) order by avg(b.score);

21、查询不同老师所教不同课程平均分从高到低显示
select t.tname, round(avg(s.score),1) from sc s,course c,teacher t 
where s.cno = c.cno and c.tno =t.tno 
group by t.tname;


22、查询如下课程成绩第 3 名到第 6 名的学生成绩单:企业管理(001),马克思(002),UML (003),数据库(004)
    [学生ID],[学生姓名],企业管理,马克思,UML,数据库,平均成绩
    select * from(
    select rownum rn, s1.sno,s1.sname,c.cname,s.score from  student s1,sc s,course c
     where rownum<=6 and c.cno =s.cno and s.cno=1 and s1.sno =s.sno
     group by rownum ,s1.sno,s1.sname,c.cname,s.score
     order by s.score desc )  t where t.rownum<=6 and t.rownum>=3
    
    select  s.sname,round(avg(b.score),1) ,
    (select score from sc t where cno =1 and c.cno=t.sno ) 计算机原理,
    (select score from sc t where cno =3 and c.cno=t.sno ) 数据结构,
    (select score from sc t where cno =5 and c.cno=t.sno ) 数据库原理,
    (select score from sc t where cno =6 and c.cno=t.sno ) Java语言
    from student s,sc b,course c
    where s.sno = b.sno and b.cno = b.cno 
    group by s.sname

select * from course;

23、统计列印各科成绩,各分数段人数:课程ID,课程名称,[100-85],[85-70],[70-60],[ <60]
 select c.cno "课程编码",
        c.cname "课程名称",
        sum(case when s.score between 85 and 100 then 1 else 0 end)  "100-85",--别名如果是数字开头的话,就要双引号
        sum(case when s.score between 70 and 85 then  1 else 0 end)  "85-70",
        sum(case when s.score between 60 and 70 then 1 else 0 end)  "70-60",
        sum(case when s.score <60 then 1 else 0 end)   "<60 "
        from sc s,course c
      where s.cno = c.cno 
      group by c.cno,c.cname;
      
select * from sc where cno=6 and score between 85 and 100;--between 较小数 and 较大数
update sc set score = 85 where sno =5 and cno =6;
                                          
24、查询学生平均成绩及其名次

 select
       s.sname 姓名,
       round(avg(t.score),1) 平均成绩
       from student s,sc t
       where s.sno = t.sno
       group by s.sname 
       order by round(avg(t.score),1) desc

25、查询各科成绩前三名的记录:(不考虑成绩并列情况)
      select s.sno,s.sname ,
        ( select top 3 score from sc where sno = s.sno )
        from student s,sc t
       where t.cno = c.cno 
       
       --使用 row_number() over(partition by  order by  ),rank(), derank();函数
          select * from (select sno,
                     cno,
                     score,
                     row_number() over(partition by cno order by score desc) rn
                     from sc s
                     )
          where rn <= 3 

26、查询每门课程被选修的学生数
     select cno,count(cno) from sc group by cno
     
     select * from sc where cno = 1;
27、查询出只选修了一门课程的全部学生的学号和姓名
     
     
28、查询男生、女生人数
    select  (select count(*) from student where ssex='女')  男生数, (select count(*) from student where ssex='男') 女生数 from dual;
    select * from student;
29、查询姓“张”的学生名单
    select * from student t where t.sname like '张%';
    
30、查询同名同性学生名单,并统计同名人数
    select s.*,count(*) from student s (select * from student) a where s.sname =a.sanme 
    select sname,count(*) from student s group by s.sname having count(*)>1;
31、1981年出生的学生名单(注:Student表中Sage列的类型是datetime)
    select * from student where sage between to_date('1981-01-01','yyyy-mm-dd') and to_date('1981-12-31','yyyy-mm-dd');
    select Sname,  CONVERT(char (11),DATEPART(year,Sage)) as age
    from student
    where  CONVERT(char(11),DATEPART(year,Sage))='1981';
32、查询每门课程的平均成绩,结果按平均成绩升序排列,平均成绩相同时,按课程号降序排列
    select s.cno ,t.cname ,round( avg(s.score),1) 
    from sc s,course t
    where s.cno = t.cno
    group by s.cno,t.cname
    order by round( avg(s.score),1) asc,cno desc
    
    select avg(score) from sc where cno =7
33、查询平均成绩大于85的所有学生的学号、姓名和平均成绩
   
    select s.sno 学号,s.sname 姓名,round(avg(t.score),1) 平均成绩
    from student s,sc t
    where s.sno = t.sno
    group by s.sno,s.sname 
    having round(avg(t.score),1)>60;
    
34、查询课程名称为“数据库”,且分数低于60的学生姓名和分数
    select s.sname,t.score 
    from student s,sc t,course c
    where 
    s.sno =t.sno
    and t.cno = c.cno
    and c.cname like '%数据库%'
    and t.score <60;

    select * from course;
35、查询所有学生的选课情况;
    select s.sno,s.sname,t.cname 
    from student s,course t,sc c
    where s.sno = c.sno and c.cno = t.cno
    order by sno

36、查询任何一门课程成绩在70分以上的姓名、课程名称和分数;
    select s.sname,c.cname,t.score 
    from student s,sc t,course c
    where s.sno =t.sno 
    and t.cno = c.cno 
    and exists (select * from sc a  where t.score >70 and a.sno= t.sno)

37、查询不及格的课程,并按课程号从大到小排列
    select distinct t.cno, c.cname 
    from sc t,course c 
    where t.cno = c.cno
    and t.score <60
    order by t.cno;
38、查询课程编号为003且课程成绩在80分以上的学生的学号和姓名;
    select s.sno,s.sname from student s,sc t
    where s.sno = t.sno 
    and t.cno = 3
    group by s.sno,s.sname
    having min(t.score)>80;
    
    select * from sc where  cno =3
    and not exists ( select * from sc a where a.score <80 and a.sno = s.sno)
    select * from sc t where  not exists( select * from sc t where t.score <80)
    
39、求选了课程的学生人数
    select count(*) from sc where 
40、查询选修“叶平”老师所授课程的学生中,成绩最高的学生姓名及其成绩
    select  s.sname,a.score
    from sc a,course c,student s,teacher t
    where a.sno = s.sno 
    and   a.cno = c.cno 
    and   t.tno = c.tno
    and   t.tname like '叶平'
    and   a.score =(select max(score) from sc where cno = a.cno)
    group by s.sname;
    
    
    select * from teacher
    select * from teacher where tname='叶平'
    select * from sc where cno =1
    select * from course where tno =1;
    
    select * from sc a,teacher t,student s
    where a.sno = s.sno 
    and   t.no = a.cno
    and   t.tname ='叶平'
41、查询各个课程及相应的选修人数
   select c.cname,t.cno,count(*) from sc t,course c 
   where c.cno = t.cno
   group by c.cname,t.cno
42、查询不同课程成绩相同的学生的学号、课程号、学生成绩
   select s.sno,a.cno, a.score,b.score from sc a,sc b,student s
   where a.score= b.score 
   and   a.cno <>b.cno and a.sno = s.sno
    
43、查询每门功成绩最好的前两名
--!!
--如果不考虑分数重复,用RANK()
select m.学号 , m.课程号 , m.成绩 from
(
  select t.* , RANK() over(partition by 课程号 order by 成绩 desc) px from aa t
) m
where px <= 2
order by m.课程号 , m.成绩 desc

select *  from
(
  select t.*,rank() over(partition by t.cno order by score desc)  px from sc t
) m
where px<=2
order by m.cno,m.score desc

--如果考虑分数重复,用DENSE_RANK()
select m.学号 , m.课程号 , m.成绩 from
(
  select t.* , DENSE_RANK() over(partition by 课程号 order by 成绩 desc) px from aa t
) m
where px <= 2
order by m.课程号 , m.成绩 desc

select m.* from 
(
select t.*,RANK() over(partition by t.cno order by t.score desc) px from sc t 
)m
where px <=2 
order by 
44、统计每门课程的学生选修人数(超过10人的课程才统计)。要求输出课程号和选修人数,查询结果按人数降序排列,查询结果按人数降序排列,若人数相同,按课程号升序排列 
select t.cno,count(t.sno) 
from sc t
/*where count(t.sno)>10*/
group by t.cno
having count(t.sno)>10
order by count(t.sno) desc,t.cno asc

select t.cno,count(cno) from sc t group by t.cno order by cno
select * from sc where cno = 5
45、检索至少选修两门课程的学生学号
select sno 
from sc
group by sno  
having count(cno)>2
select * from sc where sno = 7
46、查询全部学生都选修的课程的课程号和课程名
     select * 
     from course c
     where c.cno in (select cno from sc group by cno)

47、查询没学过“叶平”老师讲授的任一门课程的学生姓名
select Sname from Student where Sno not in (select Sno from Course,Teacher,SC where Course.Tno=Teacher.Tno and SC.Cno=course.Cno and Tname='叶平');

48、查询两门以上不及格课程的同学的学号及其平均成绩
  select sno,avg(nvl(score,0)) from SC where sno in (select sno from SC where score <60 group by sno having count(*)>2)group by sno;

49、检索“004”课程分数小于60,按分数降序排列的同学学号
  select sno from SC where cno='004'and score <60 order by score desc;
50、删除“002”同学的“001”课程的成绩

delete from Sc where S#='002'and C#='001';
--sysdba 登录
grant select on v$statname to scott;
grant select on v$sesstat to scott;
grant select on v$mystat to scott;
grant select on v$mystat to scott;

数据库面试必知必会
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原文地址:https://www.cnblogs.com/humi/p/7569587.html