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(10) 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(10) not null, depart varchar(10) not null ); insert into student 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 course values ("3-105","计算机导论",825), ("3-245","操作系统",825), ("6-166","数据电路",825), ("9-888","高等数学",100); insert into score values (103,"3-245",86), (105,"3-245",75), (109,"3-245",68), (103,"3-245",92), (105,"3-105",88), (109,"3-105",76), (101,"3-105",64), (107,"3-105",91), (108,"3-105",78), (101,"6-166",85), (107,"6-106",79), (108,"6-166",81); insert into teacher 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"); 题目： 1、 查询Student表中的所有记录的Sname、Ssex和Class列。 2、 查询教师所有的单位即不重复的Depart列。 3、 查询Student表的所有记录。 4、 查询Score表中成绩在60到80之间的所有记录。 5、 查询Score表中成绩为85，86或88的记录。 6、 查询Student表中“95031”班或性别为“女”的同学记录。 7、 以Class降序查询Student表的所有记录。 8、 以Cno升序、Degree降序查询Score表的所有记录。 9、 查询“95031”班的学生人数。 10、查询Score表中的最高分的学生学号和课程号。 11、查询‘3-105’号课程的平均分。 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。 13、查询最低分大于70，最高分小于90的Sno列。 答案： 1、select sname,ssex,class from student; 2、select distinct depart from teacher; 3/ select * from student; 4/ select * from score where degree between 60 and 80; 5/ select * from score where degree in(85,86,88); 6/ select * from student where class="95031" or ssex = "女"; 7、 select * from student where order by class desc; 8/ select * from student where order by cno asc,degree desc; 9/ select sum(class="95031") from student; 第二中写法 SELECT COUNT(*) FROM STUDENT WHERE CLASS='95031'; 10/ select max(degree),min(degree) from score; select sno,cno from score; 最后结果为： select sno,cno from score where degree=(select max(degree),min(degree) from score); 11、 select avg(degree) from score where cno = "3-105"; 12/ select cno,avg(degree) from score where cno like "3%" groupd by cno having count(sno) >= 5; 13/ select sno from score group by sno having min(degree) > 70 and max(degree) < 90;
14、查询所有学生的Sname、Cno和Degree列。 15、查询所有学生的Sno、Cname和Degree列。 16、查询所有学生的Sname、Cname和Degree列。 17、查询“95033”班所选课程的平均分。 18、假设使用如下命令建立了一个grade表： create table grade(low number(3,0),upp number(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’); commit; 现查询所有同学的Sno、Cno和rank列。 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。 20、查询score中选学一门以上课程的同学中分数为非最高分成绩的记录。 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。 22、查询和学号为108的同学同年出生的所有学生的Sno、Sname和Sbirthday列。 23、查询“张旭“教师任课的学生成绩。 24、查询选修某课程的同学人数多于5人的教师姓名。 25、查询95033班和95031班全体学生的记录。 26、查询存在有85分以上成绩的课程Cno. 27、查询出“计算机系“教师所教课程的成绩表。 28、查询“计算机系”与“电子工程系“不同职称的教师的Tname和Prof。 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Cno、Sno和Degree,并按Degree从高到低次序排序。 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Cno、Sno和Degree. 31、查询所有教师和同学的name、sex和birthday. 32、查询所有“女”教师和“女”同学的name、sex和birthday. 33、查询成绩比该课程平均成绩低的同学的成绩表。 34、查询所有任课教师的Tname和Depart. 35 查询所有未讲课的教师的Tname和Depart. 36、查询至少有2名男生的班号。 37、查询Student表中不姓“王”的同学记录。 38、查询Student表中每个学生的姓名和年龄。 39、查询Student表中最大和最小的Sbirthday日期值。 40、以班号和年龄从大到小的顺序查询Student表中的全部记录。 41、查询“男”教师及其所上的课程。 42、查询最高分同学的Sno、Cno和Degree列。 43、查询和“李军”同性别的所有同学的Sname. 44、查询和“李军”同性别并同班的同学Sname. 45、查询所有选修“计算机导论”课程的“男”同学的成绩表 答案： 14.SELECT A.SNAME,B.CNO,B.DEGREE FROM STUDENT AS A JOIN SCORE AS B ON A.SNO=B.SNO; 15.SELECT A.CNAME, B.SNO,B.DEGREE FROM COURSE AS A JOIN SCORE AS B ON A.CNO=B.CNO ; 16.SELECT A.SNAME,B.CNAME,C.DEGREE FROM STUDENT A JOIN (COURSE B,SCORE C) ON A.SNO=C.SNO AND B.CNO =C.CNO; 17.SELECT AVG(A.DEGREE) FROM SCORE A JOIN STUDENT B ON A.SNO = B.SNO WHERE B.CLASS='95033'; 18.SELECT A.SNO,A.CNO,B.RANK FROM SCORE A,GRADE B WHERE A.DEGREE BETWEEN B.LOW AND B.UPP ORDER BY RANK; 19.SELECT A.* FROM SCORE A JOIN SCORE B WHERE A.CNO='3-105' AND A.DEGREE>B.DEGREE AND B.SNO='109' AND B.CNO='3-105'; 另一解法：SELECT A.* FROM SCORE A WHERE A.CNO='3-105' AND A.DEGREE>ALL(SELECT DEGREE FROM SCORE B WHERE B.SNO='109' AND B.CNO='3-105'); 20.SELECT * FROM score s WHERE DEGREE<(SELECT MAX(DEGREE) FROM SCORE) GROUP BY SNO HAVING COUNT(SNO)>1 ORDER BY DEGREE ; 21.见19的第二种解法 22。SELECT SNO,SNAME,SBIRTHDAY FROM STUDENT WHERE YEAR(SBIRTHDAY)=(SELECT YEAR(SBIRTHDAY) FROM STUDENT WHERE SNO='108'); ORACLE:select x.cno,x.Sno,x.degree from score x,score y where x.degree>y.degree and y.sno='109'and y.cno='3-105'; select cno,sno,degree from score where degree >(select degree from score where sno='109' and cno='3-105') 23.SELECT A.SNO,A.DEGREE FROM SCORE A JOIN (TEACHER B,COURSE C) ON A.CNO=C.CNO AND B.TNO=C.TNO WHERE B.TNAME='张旭'; 另一种解法：select cno,sno,degree from score where cno=(select x.cno from course x,teacher y where x.tno=y.tno and y.tname='张旭'); 根据实际EXPLAIN此SELECT语句，第一个的扫描次数要小于第二个 24.SELECT A.TNAME FROM TEACHER A JOIN (COURSE B, SCORE C) ON (A.TNO=B.TNO AND B.CNO=C.CNO) GROUP BY C.CNO HAVING COUNT(C.CNO)>5; 另一种解法：select tname from teacher where tno in(select x.tno from course x,score y where x.cno=y.cno group by x.tno having count(x.tno)>5); 实际测试1明显优于2 25。select cno,sno,degree from score where cno=(select x.cno from course x,teacher y where x.tno=y.tno and y.tname='张旭'); 26。SELECT CNO FROM SCORE GROUP BY CNO HAVING MAX(DEGREE)>85; 另一种解法：select distinct cno from score where degree in (select degree from score where degree>85); 27。SELECT A.* FROM SCORE A JOIN (TEACHER B,COURSE C) ON A.CNO=C.CNO AND B.TNO=C.TNO WHERE B.DEPART='计算机系'; 另一种解法：SELECT * from score where cno in (select a.cno from course a join teacher b on a.tno=b.tno and b.depart='计算机系'); 此时2略好于1，在多连接的境况下性能会迅速下降 28。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'); 29。SELECT * FROM SCORE WHERE DEGREE>ANY(SELECT DEGREE FROM SCORE WHERE CNO='3-245') ORDER BY DEGREE DESC; 30。SELECT * FROM SCORE WHERE DEGREE>ALL(SELECT DEGREE FROM SCORE WHERE CNO='3-245') ORDER BY DEGREE DESC; 31.SELECT SNAME AS NAME, SSEX AS SEX, SBIRTHDAY AS BIRTHDAY FROM STUDENT UNION SELECT TNAME AS NAME, TSEX AS SEX, TBIRTHDAY AS BIRTHDAY FROM TEACHER; 32.SELECT SNAME AS NAME, SSEX AS SEX, SBIRTHDAY AS BIRTHDAY FROM STUDENT WHERE SSEX='女' UNION SELECT TNAME AS NAME, TSEX AS SEX, TBIRTHDAY AS BIRTHDAY FROM TEACHER WHERE TSEX='女'; 33.SELECT A.* FROM SCORE A WHERE DEGREE<(SELECT AVG(DEGREE) FROM SCORE B WHERE A.CNO=B.CNO); 须注意********此题 34。解法一：SELECT A.TNAME,A.DEPART FROM TEACHER A JOIN COURSE B ON A.TNO=B.TNO; 解法二：select tname,depart from teacher a where exists (select * from course b where a.tno=b.tno); 解法三：SELECT TNAME,DEPART FROM TEACHER WHERE TNO IN (SELECT TNO FROM COURSE); 实际分析，第一种揭发貌似更好，至少扫描次数最少。 35.解法一：SELECT TNAME,DEPART FROM TEACHER A LEFT JOIN COURSE B USING(TNO) WHERE ISNUL (B.tno); 解法二：select tname,depart from teacher a where not exists (select * from course b where a.tno=b.tno); 解法三：SELECT TNAME,DEPART FROM TEACHER WHERE TNO NOT IN (SELECT TNO FROM COURSE); NOT IN的方法效率最差，其余两种差不多 36.SELECT CLASS FROM STUDENT A WHERE SSEX='男' GROUP BY CLASS HAVING COUNT(SSEX)>1; 37.SELECT * FROM STUDENT A WHERE SNAME not like '王%'; 38.SELECT SNAME,(YEAR(NOW())-YEAR(SBIRTHDAY)) AS AGE FROM STUDENT; 39.select sname,sbirthday as THEMAX from student where sbirthday =(select min(SBIRTHDAY) from student) union select sname,sbirthday as THEMIN from student where sbirthday =(select max(SBIRTHDAY) from student); 40.SELECT CLASS,(YEAR(NOW())-YEAR(SBIRTHDAY)) AS AGE FROM STUDENT ORDER BY CLASS DESC,AGE DESC; 41.SELECT A.TNAME,B.CNAME FROM TEACHER A JOIN COURSE B USING(TNO) WHERE A.TSEX='男'; 42.SELECT A.* FROM SCORE A WHERE DEGREE=(SELECT MAX(DEGREE) FROM SCORE B ); 43.SELECT SNAME FROM STUDENT A WHERE SSEX=(SELECT SSEX FROM STUDENT B WHERE B.SNAME='李军'); 44.SELECT SNAME FROM STUDENT A WHERE SSEX=(SELECT SSEX FROM STUDENT B WHERE B.SNAME='李军' ) AND CLASS=(SELECT CLASS FROM STUDENT C WHERE c.SNAME='李军'); 45.解法一：SELECT A.* FROM SCORE A JOIN (STUDENT B,COURSE C) USING(sno,CNO) WHERE B.SSEX='男 ' AND C.CNAME='计算机导论'; 解法二：select * from score where sno in(select sno from student where ssex='男') and cno=(select cno from course where cname='计算机导论');
emp员工表 字段如下： empno 员工号 ename 员工姓名 job 职位 mgr 上级编号 hiredate入职日期 sal 薪资 comm 提成 deptno 部门编号 --1、选择部门30中的员工； select *from emp where deptno=30; --2、查询所有办事员（CLERK）的姓名、编号和部门； select ename,empno,deptno from emp where job ='CLERK'; --3、找出提成高于工资的员工； select * from emp where comm>sal; --4、找出提成高于薪资60%的员工 select * from emp where comm>sal*0.6; --5、找出部门10中所有经理和部门20中所有办事员的详细资料 select * from emp where deptno=10 and job='MANAGER' or deptno=20 and job ='CLERK' ; ============================= --6、找出部门10中所有经理、部门20中所有办事员以及既不是经理又不是办事员但薪资大于或等于2000的所有雇员的详细资料 select * from emp where deptno=10 and job='MANAGER' or deptno=20 and job ='CLERK' or sal>=2000 and job not in('MANAGER','CLERK'); 第二种方法 （换 where后面第二个or） or (job !='MANAGER' and job!='CLERK'and sal>=1200); --7、找出收取提成的员工的不同职位； select distinct job from emp where comm is not null; --8、找出不收取提成 或收取提成低于100的员工； select * from emp where comm is null or comm<100; --9、显示雇员的详细资料，按姓名排序 select * from emp where job!='MANAGER'and job!='CLERK' order by ename asc; --10、显示员工姓名，根据工作年限将最老的员工排在最前面 select ename,hiredate from emp order by hiredate asc; --11、显示所有员工的姓名、部门编号和薪资，部门编号降序排列，同部门员工以薪资升序排列 select ename,deptno,sal,job from emp order by deptno desc , sal; --12、显示姓名字段的任何位置都包含A的所有员工姓名 select ename from emp where ename like '%A%';