• 数据库45道练习题


    1、 查询Student表中的所有记录的Sname、Ssex和Class列。

    SELECT sname,ssex,class FROM student;

    2、 查询教师所有的单位即不重复的Depart列。distinct

    SELECT DISTINCT depart FROM teacher;

    3、 查询Student表的所有记录。

    SELECT * FROM student;

    4、 查询Score表中成绩在60到80之间的所有记录。between and

    SELECT *FROM score WHERE degree BETWEEN 60 AND 80;

    5、查询Score表中成绩为85,86或88的记录

    解法1:select * FROM score WHERE degree IN(85,86,88);
    解法2:select * FROM score WHERE degree = 85 OR degree =86 OR degree =88;

    6、 查询Student表中“95031”班或性别为“女”的同学记录。

    select * FROM student WHERE class = '95031' OR ssex = '';

    7、 以Class降序查询Student表的所有记录。order BY

    select * FROM student ORDER BY class DESC;

    8、 以Cno升序、Degree降序查询Score表的所有记录。

    select * FROM score ORDER BY cno,degree DESC;

    9、 查询“95031”班的学生人数。

    SELECT * FROM student WHERE class = '95031';

    10、查询Score表中的最高分的学生学号和课程号。(子查询或者排序)

    SELECT sno,cno,degree FROM score WHERE degree = (SELECT MAX(degree) FROM score);子查询
    SELECT sno,cno,degree FROM score ORDER BY degree DESC LIMIT 1;排列

    11、查询每门课的平均成绩。

    SELECT cno,AVG(degree) FROM score GROUP BY cno;

    12、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。

    SELECT cno,AVG(degree) FROM score GROUP BY cno HAVING COUNT(*)>=5 AND cno LIKE '3%';

    13、查询分数大于70,小于90的Sno列。

    SELECT sno FROM score WHERE degree >70 AND degree <90;

    14、查询所有学生的Sname、Cno和Degree列。

    select aa.sname,bb.cno,bb.degree FROM student AS aa,score AS bb WHERE aa.`sno` = bb.`sno`;

    15、查询所有学生的Sno、Cname和Degree列。

    解法1:
    SELECT a.sname,b.degree,c.cname FROM student AS a,score AS b,course AS c WHERE a.sno = b.sno AND b.cno = c.cno;
    解法2:
    SELECT a.sname,b.degree,c.cname FROM student AS a JOIN score AS b ON a.sno = b.sno JOIN course AS c ON b.cno = c.cno; 

    16、 查询所有学生的Sname、Cname和Degree列。(三表联查)

    SELECT a.sno,b.degree,c.cname FROM student AS a JOIN score AS b ON a.sno = b.sno JOIN course AS c ON b.cno = c.cno; 

    17、 查询“95031”班学生的平均分。

    SELECT AVG(degree) FROM score WHERE sno = ANY(SELECT sno FROM student WHERE class = '95031');

     18、假设使用如下命令建立了一个grade表:

    create table grade(low  int(3),upp  int(3),rank  char(1))

    insert into grade values(90,100,’A’)

    insert into grade values(80,89,’B’)

    insert into grade values(70,79,’C’)

    insert into grade values(60,69,’D’)

    insert into grade values(0,59,’E’)

    现查询所有同学的SnoCnorank列。

    解法1:select sno,cno,rank from score,grade where degree between low and upp order by rank;
    解法2:select sno,cno,degree,[rank]
    from Score join grade on degree between low and upp order by [rank]--on后面加筛选条件
    解法3:select Sno,Cno,(select [rank] from grade where Score.Degree between low and upp) as LV
    from Score order by Degree desc--子查询得到的结果必须唯一

    19查询选修“3-105”课程的成绩高于“109”号同学成绩的所有同学的记录。

    select * from score where cno='3-105' and degree>(select degree from score  where sno=109 and cno='3-105');

    20、查询score中选学多门课程的同学中分数为非最高分成绩的记录。

    理解1:选学多门课程的同学中,分数小于所有学生中最高分:
    1.select * from score where sno in2and degree<32.select sno from score group by sno having count(*)>1
    3.select max(degree) from score  where sno in44.select sno from score group by sno having count(*)>1
    结合起来:select * from score where sno in(select sno from score group by sno having count(*)>1) and
    degree<(select max(degree) from score  where sno in(select sno from score group by sno having count(*)>1));
    理解2:选多门课程中,分数小于每门课程最高分的学生信息
    select * from score a where sno in(select sno from score group by sno having count(*)>1) and
    degree<(select max(degree) from score b where b.cno = a.cno )

    21 查询成绩高于学号为“109”、课程号为“3-105”的成绩的所有记录。

    select *from score where degree>(select degree from score where sno='109'and  Cno='3-105')

    22、查询和学号为108的同学同年出生的所有学生的SnoSnameSbirthday

    select sno,sname,sbirthday from student where year(sbirthday)=(select year(sbirthday) from student where sno='108');

    23、查询张旭教师任课的学生成绩。

    1.select * from score where cno in ()
    2.select cno from course where tno=()
    3.select tno from teacher where tname='张旭'
    结合起来:
    select * from score where cno in (select cno from course where tno=(select tno from teacher where tname='张旭'))
    解法1:select teacher.tname,course.cno,score.degree from teacher,course,score where teacher.tno=course.tno and course.cno=score.cno and tname='张旭';
    解法2:select degree from Score join Course on Score.Cno=Course.Cno join Teacher on Course.Tno=Teacher.Tno where Tname = '张旭'

    24、查询选修某课程的同学人数多于5人的教师姓名。

    1.select tname from teacher where tno=()
    2.select tno from course where cno=()
    3.select cno from score group by cno having count(*)>5
    联合起来:
    select tname from teacher where tno=(select tno from course where cno=(select cno from score group by cno having count(*)>5))
    解法1;select teacher.tname from teacher,score,course where teacher.tno=course.tno and course.cno=score.cno group by tname having count(*)>5 ;
    解法2:select Tname from Teacher where Tno=(select Tno from Course where Cno=(select Cno from Score group by Cno having COUNT(Cno)>=5))

    25、查询95033班和95031班全体学生的记录。

    解法1:select * from student where class='95033' or class='95031';
    解法2:select*from student inner join Score on Student.Sno=Score.Sno where Class in(‘95033’,‘95031’)
    解法3:select *from Student,Score where Class in(95033,95031) and Student.Sno=Score.Sno

    26、 查询存在有85分以上成绩的课程Cno.

    select score.cno from score,course where score.cno=course.cno and degree>85;
    select cno from score group by cno having max(degree)>85;

    27、查询出计算机系教师所教课程的成绩表

    解法1:select degree from teacher,score,course where teacher.tno=course.tno and course.cno=score.cno and teacher.depart='计算机系';
    解法2:select *from Score where Cno in( select Cno from Course where Tno in (select Tno from Teacher where Depart='计算机系'))
    解法3:select Sno,Score.Cno,Degree from Score join Course on Score.Cno=Course.Cno join Teacher on Course.Tno=Teacher.Tno where Depart='计算机系'

    28、查询计算 机系电子工程系不同职称的教师的TnameProf

    解法1:select tname,prof from teacher where depart='计算机系' and prof not in (select prof from teacher where depart='电子工程系');
    解法2:select tname,prof from teacher where prof not in
    (select prof from teacher where depart='电子工程系' and prof in 
    (select prof from teacher where depart='计算机系'))and
    depart in ('计算机系','电子工程系')--查询两个系中教师相同职称名称,不在这里面的就是除去两个系都有的剩下的,
    见上面select prof from teacher where depart='电子工程系' and prof in (select prof from teacher where depart='计算机系')

    29、查询选修编号为“3-105“课程且成绩至少高于选修编号为“3-245”的同学的CnoSnoDegree,并按Degree从高到低次序排序。any的用法

    解法1;select * from score where cno='3-105' and degree > (select min(degree) from score where cno='3-245')order by degree desc;
    解法2:select * from score where cno='3-105' and degree > any(select degree from score where cno='3-245')order by degree desc;

    30、查询选修编号为“3-105”且成绩高于选修编号为“3-245”课程的同学的CnoSnoDegree.all的用法)

    解法1:select * from score where cno='3-105' and degree > (select max(degree) from score where cno='3-245');
    解法2:select * from score where cno='3-105' and degree > all(select degree from score where cno='3-245');

    31 查询所有教师和同学的namesexbirthday.

    select tname,tsex,tbirthday from teacherunionselect sname,ssex,sbirthday from student

    32、查询所有教师和同学的namesexbirthday.

    select tname,tsex,tbirthday from teacher where tsex=''unionselect sname,ssex,sbirthday from student where ssex=''

    33 查询成绩比该课程平均成绩低的同学的成绩表。

    select * from score as aa where degree<(select avg(degree) from score as bb where aa.cno=bb.cno);
    解法2:详解--相关子查询--同一门学科的平均分,,每门学科低于自身平均分的
    select * from score as a where a.degree<
    (select AVG(degree) from score as b where a.cno = b.cno group by cno)--相当于foreach列出每一门课的平均分select * from score as a where a.degree< (select AVG(degree) from score as b group by cno having a.cno = b.cno)

    34、 查询所有任课教师的TnameDepart

    解法1:select tname,depart from teacher where exists (select * from course where teacher.tno=course.tno);
    解法2:select Tname,Depart from Teacher where Tno in(select distinct Tno from Course)  

    35 、 查询所有未讲课的教师的TnameDepart.

    解法1:select tname,depart from teacher where not exists (select * from course where teacher.tno=course.tno);
    解法2:select Tname,Depart from Teacher where Tno in
    (select Tno from Course where Cno not in(select Cno from Score group by Cno))
    解法3:select Tname,Depart from Teacher where Tno not in(select Tno from Course)

    36、查询至少有2名男生的班号。

    select class from student where ssex=''group by class having count(*)>=2;

    37、查询Student表中不姓的同学记录。

    select * from student where sname not like'王%'; 

    38、查询Student表中每个学生的姓名和年龄。

    select sname,year(now())-year(sbirthday) from student; 

    39、查询Student表中最大和最小的Sbirthday日期值。

    select max( date(student.sbirthday)) ,min( date(student.sbirthday)) from student; 

    40、以班号和年龄从大到小的顺序查询Student表中的全部记录。

    select * from student order by class desc,date(sbirthday) asc; 

    41、查询教师及其所上的课程。

    解法1:select teacher.tname,teacher.tsex,course.cname from teacher,course where teacher.tno=course.tno and tsex=''; 
    解法2:select teacher.tno,tname,tsex,cname,cno from teacher join course on teacher.tno=course.tno where tsex=''

    42、查询最高分同学的SnoCnoDegree列。

    select sno,cno,degree from score where degree=(select max(degree) from score);

    43、查询和李军同性别的所有同学的Sname.

    select sname from student where ssex=(select ssex from student where sname='李军');

    44、查询和李军同性别并同班的同学Sname.

    select sname from student where ssex=(select ssex from student where sname='李军') and class=(select class from student where sname='李军');

    45、查询所有选修计算机导论课程的同学的成绩表。

    select student.sname,score.degree from student,score,course 
    where student.sno=score.sno and course.cno=score.cno and course.cname='计算机导论' and student.ssex='';
    select * from score where sno in(select sno from student where ssex='') and cno=(select cno from course where cname='计算机导论')
  • 相关阅读:
    java oop第09章_JDBC02(CRUD操作)
    java 与日期转换相关的方法(java.util.date类型和java.sql.date类型互相转换)、随机字符串生成方法、UUID生产随机字符串
    Java oop第08章_JDBC01(入门)
    java 数组中的数值反转输出
    java oop第07章_集合框架
    Java oop创建自定义异常
    java oop遍历List和Map的几种方法
    java oop第06章_异常处理
    Java oop第05章_多态、接口
    信开发 新浪SAE开发平台 验证Token 一直失败
  • 原文地址:https://www.cnblogs.com/mr171733/p/9597482.html
Copyright © 2020-2023  润新知