答案如下--创建并使用数据库------------*/ create database 50q; use 50q; /*-------------------建表-------------------------*/ CREATE TABLE STUDENT ( SNO VARCHAR(3) NOT NULL, SNAME VARCHAR(4) NOT NULL, SSEX VARCHAR(2) NOT NULL, SBIRTHDAY DATETIME, CLASS VARCHAR(5) ); CREATE TABLE COURSE ( CNO VARCHAR(5) NOT NULL, CNAME VARCHAR(10) NOT NULL, TNO VARCHAR(3) NOT NULL ); CREATE TABLE SCORE ( SNO VARCHAR(3) NOT NULL, CNO VARCHAR(5) NOT NULL, DEGREE NUMERIC(10, 1) NOT NULL ) ; CREATE TABLE TEACHER ( TNO VARCHAR(3) NOT NULL, TNAME VARCHAR(4) NOT NULL, TSEX VARCHAR(2) NOT NULL, TBIRTHDAY DATETIME NOT NULL, PROF VARCHAR(6), DEPART VARCHAR(10) NOT NULL ); /*---------------添加主键--------------*/ ALTER TABLE STUDENT ADD PRIMARY KEY (SNO); ALTER TABLE SCORE ADD PRIMARY KEY (SNO,CNO); ALTER TABLE COURSE ADD PRIMARY KEY (CNO); ALTER TABLE TEACHER ADD PRIMARY KEY (TNO); #--------主键在两张table中的数据类型须一致-------*/ ALTER TABLE SCORE ADD CONSTRAINT FK_SCORE_STUDENT FOREIGN KEY (SNO) REFERENCES STUDENT(SNO); ALTER TABLE SCORE ADD CONSTRAINT FK_SCORE_COURSE FOREIGN KEY (CNO) REFERENCES COURSE(CNO); ALTER TABLE COURSE ADD CONSTRAINT FK_COURSE_TEACHER FOREIGN KEY (TNO) REFERENCES TEACHER(TNO); /*---------------输入记录-------------------*/ INSERT INTO STUDENT (SNO,SNAME,SSEX,SBIRTHDAY,CLASS) VALUES (108 ,'曾华' ,'男' ,'1977-09-01',95033), (105 ,'匡明' ,'男' ,'1975-10-02',95031), (107 ,'王丽' ,'女' ,'1976-01-23',95033), (101 ,'李军' ,'男' ,'1976-02-20',95033), (109 ,'王芳' ,'女' ,'1975-02-10',95031), (103 ,'陆君' ,'男' ,'1974-06-03',95031); INSERT INTO TEACHER(TNO,TNAME,TSEX,TBIRTHDAY,PROF,DEPART) VALUES (804,'李诚','男','1958-12-02','副教授','计算机系'), (856,'张旭','男','1969-03-12','讲师','电子工程系'), (825,'王萍','女','1972-05-05','助教','计算机系'), (831,'刘冰','女','1977-08-14','助教','电子工程系'); INSERT INTO COURSE(CNO,CNAME,TNO) VALUES ('3-105' ,'计算机导论',825), ('3-245' ,'操作系统' ,804), ('6-166' ,'数据电路' ,856), ('9-888' ,'高等数学' ,831); INSERT INTO SCORE(SNO,CNO,DEGREE) VALUES (103,'3-245',86),(105,'3-245',75), (109,'3-245',68),(103,'3-105',92), (105,'3-105',58),(109,'3-105',46), (101,'3-105',64),(107,'3-105',91), (108,'3-105',78),(101,'6-166',85), (107,'6-166',79),(108,'6-166',81);
insert into score (sno,cno,degree) value('101','9-888','50');
update score set degree=60 where cno='3-105' and sno='105';
select * from score where degree between 60 and 80 order by cno ,degree desc;
select sno,avg(DEGREE) as avger from score group by sno having avg(DEGREE)>60;
select student.sno, student.sname from student inner join score on student.sno = score.sno group by sno having count(cno)=2;
select count(tname) from teacher where tname like '张%';
select cno,max(degree),min(degree) from score group by cno;
select distinct score.sno,sname from score inner join student on score.sno=student.sno where cno in (select cno from score where sno='101');
select student.sno,student.sname from student,teacher where teacher.tname='王萍';
select Class FROM student where Ssex='男' group by Class having COUNT(*)>1;
CREATE DEFINER=`root`@`localhost` PROCEDURE `pp5`(in cn varchar(5),out av int,out ma int) BEGIN select round(avg(degree)),round(max(degree)) into av,ma from score where cno=cn; END
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