常见的sql语句练习

标签：str   ash   问题：   sse   合并   修改   lte   厦门   play   

一、

1.新建表 test

id varchar2(20)
name varchar2(20)
addr varchar2(50)
score number

create table test
(id varchar2(20),
name varchar2(20),
addr varchar2(50),
score number
)

2.给id增加主键约束
alter table test add constraint pk_id primary key(id)
3.查询出分数大于60分的人员，并按分数降序进行排列

select * from test where score>60 order by score desc
4.新增数据 001 zhangsan shenzhen 80
insert into test(id,name,addr,score) values (‘001‘,‘zhansgan‘,‘shenzhen‘,‘80‘)

6. 将zhangsan的分数修改成60
update test set score=60 where name=‘zhangsan‘

7.删除zhangsan的信息

delete from test where name=‘zhangsan‘

二、

Student(S#,Sname,Sage,Ssex) 学生表
Course(C#,Cname,T#) 课程表
SC(S#,C#,score) 成绩表
Teacher(T#,Tname) 教师表

问题：

1、查询平均成绩大于60分的同学的学号和平均成绩； avg

select s#,avg(score) from sc where (select avg(score) from sc)>60 group by s# ;

select s#,avg(score) from sc where group by s# having avg(score)>60;
2、查询所有同学的学号、姓名、选课数、总成绩；

select Student.s#,sname,count(C#),sum(score) from Student inner join sc on Student.s#=sc.s# group by Student.s#,sname

4、查询没学过“叶平”老师课的同学的学号、姓名；

select S#,Sname from Student where s# in (select S# from sc where c# not in (select c# from Course where T#=(select T# from teacher where Tname=‘叶平‘ ))

5、查询所有课程成绩小于60分的同学的学号、姓名；

select S#,Sname from Student where s# not in (select S# from sc where score>60)

———————————————————————————————————————
向student表插入记录的INSERT语句如下：

INSERT INTO student VALUES( 901,‘张老大‘, ‘男‘,1985,‘计算机系‘, ‘北京市海淀区‘);

INSERT INTO student VALUES( 902,‘张老二‘, ‘男‘,1986,‘中文系‘, ‘北京市昌平区‘);

INSERT INTO student VALUES( 903,‘张三‘, ‘女‘,1990,‘中文系‘, ‘湖南省永州市‘);

INSERT INTO student VALUES( 904,‘李四‘, ‘男‘,1990,‘英语系‘, ‘辽宁省阜新市‘);

INSERT INTO student VALUES( 905,‘王五‘, ‘女‘,1991,‘英语系‘, ‘福建省厦门市‘);

INSERT INTO student VALUES( 906,‘王六‘, ‘男‘,1988,‘计算机系‘, ‘湖南省衡阳市‘);

———————————————————————————————————————

6.从student表中查询年龄18~22岁的学生信息

SELECT id,name,sex,2010-birth AS age,department,address FROM student WHERE 2010-birth>=18 AND 2010-birth<=22;     （其实没有age，直接计算今年年份-出生birth） 

SELECT id,name,sex,2013-birth AS age,department,address FROM student WHERE 2013-birth BETWEEN  18 AND 22;

7.从student表中查询每个院系有多少人

SELECT department, COUNT(id) FROM student GROUP BY department;

8.从score表中查询每个科目的最高分

SELECT c_name,MAX(grade) FROM score GROUP BY c_name;

9.查询李四的考试科目（c_name）和考试成绩（grade）

SELECT c_name, grade FROM score WHERE stu_id=(SELECT id FROM student WHERE name= ‘李四‘ );

10.用连接的方式查询所有学生的信息和考试信息

SELECT student.id,name,sex,birth,department,address,c_name,grade FROM student,score WHERE student.id=score.stu_id;

11.计算每个学生的总成绩

SELECT student.id,name,SUM(grade) FROM student,score WHERE student.id=score.stu_id GROUP BY id; (两个表都有id，要注意冲突)

12.计算每个考试科目的平均成绩

SELECT c_name,AVG(grade) FROM score GROUP BY c_name;

13.查询计算机成绩低于95的学生信息

SELECT * FROM student WHERE id IN (SELECT stu_id FROM score WHERE c_name="计算机" and grade<95);

14.查询同时参加计算机和英语考试的学生的信息

SELECT * FROM student WHERE id =ANY ( SELECT stu_id FROM score WHERE stu_id IN ( SELECT stu_id FROM score WHERE c_name= ‘计算机‘) AND c_name= ‘英语‘ );（=ANY 与子查询IN相同）

SELECT a.* FROM student a ,score b ,score c WHERE a.id=b.stu_id  AND b.c_name=‘计算机‘ AND a.id=c.stu_id AND c.c_name=‘英语‘;

15.将计算机考试成绩按从高到低进行排序

SELECT stu_id, grade FROM score WHERE c_name= ‘计算机‘ ORDER BY grade DESC;

16.从student表和score表中查询出学生的学号，然后合并查询结果

SELECT id FROM student UNION SELECT stu_id FROM score;

17.查询都是湖南的学生的姓名、年龄、院系和考试科目及成绩

 SELECT student.id, name,sex,birth,department, address, c_name,grade FROM student, score WHERE address LIKE ‘湖南%‘ AND student.id=score.stu_id;

18.查询姓张或者姓王的同学的姓名、院系和考试科目及成绩

SELECT student.id, name,sex,birth,department, address, c_name,grade FROM student, score WHERE (name LIKE ‘张%‘ OR name LIKE ‘王%‘) AND student.id=score.stu_id ;

19.查询student表中学生的学号、姓名、年龄、院系和籍贯并且按照年龄从小到大的顺序排列。
select student.id,name,2017-birth,department,address from student ORDER BY 2017-birth；

20.统计超过2人考试的科目

select c_name ,count(*) from score group by c_name having count(*)>2

常见的sql语句练习

标签：str   ash   问题：   sse   合并   修改   lte   厦门   play   

原文地址：https://www.cnblogs.com/hjy123/p/12160744.html