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最强SQL思维举重,完爆所有sql笔试题

数据库 yangshuai 63702浏览 0评论

你有没有感觉笔试的时间不够用,你有没有因为sql题目浪费太多时间,你没有对sql题目感到很触头。来吧,把下面的题目好好练习一下,保证你对sql的理解进入一个新的层次,对战笔试题更是战无不胜,攻无不克。

建表和初始化sql:

–创建表
create table T_STUDENT(sno NUMBER not null, sname VARCHAR2(30), sdree VARCHAR2(50), sage NUMBER, ssex CHAR(2));
alter table T_STUDENT add primary key (SNO);
create table T_SCORE(sno NUMBER, cno NUMBER, grade NUMBER(4,1), tno NUMBER, id NUMBER not null);
alter table T_SCORE add primary key (ID);
create table T_COURSE(cno NUMBER not null, cname VARCHAR2(30));
alter table T_COURSE add primary key (CNO);

–初始化学生表
insert into T_STUDENT (SNO, SNAME, SDREE, SAGE, SSEX) values (1, ‘李坤’, ‘天融信’, 26, ‘男’);
insert into T_STUDENT (SNO, SNAME, SDREE, SAGE, SSEX) values (2, ‘曹贵生’, ‘中银’, 26, ‘男’);
insert into T_STUDENT (SNO, SNAME, SDREE, SAGE, SSEX) values (3, ‘柳波’, ‘淘宝’, 27, ‘男’);
insert into T_STUDENT (SNO, SNAME, SDREE, SAGE, SSEX) values (4, ‘纪争光’, ‘IBM’, 23, ‘男’);
insert into T_STUDENT (SNO, SNAME, SDREE, SAGE, SSEX) values (5, ‘李学宇’, ‘微软’, 25, ‘女’);
insert into T_STUDENT (SNO, SNAME, SDREE, SAGE, SSEX) values (6, ‘李雪琪’, ‘文思’, 25, ‘女’);
insert into T_STUDENT (SNO, SNAME, SDREE, SAGE, SSEX) values (7, ‘陈绪’, ‘百度’, 26, ‘男’);
insert into T_STUDENT (SNO, SNAME, SDREE, SAGE, SSEX) values (8, ‘韩正阳’, ‘中海油’, 24, ‘男’);
insert into T_STUDENT (SNO, SNAME, SDREE, SAGE, SSEX) values (9, ‘陈伟东’, ‘腾讯’, 24, ‘男’);
insert into T_STUDENT (SNO, SNAME, SDREE, SAGE, SSEX) values (10, ‘刘兵’, ‘华为’, 24, ‘男’);
insert into T_STUDENT (SNO, SNAME, SDREE, SAGE, SSEX) values (11, ‘丁成云’, ‘联想’, 25, ‘女’);
insert into T_STUDENT (SNO, SNAME, SDREE, SAGE, SSEX) values (12, ‘王鹏’, ‘中兴’, 25, ‘男’);
commit;

–初始化课程表
insert into T_COURSE (CNO, CNAME) values (1, ‘JAVA程序设计’);
insert into T_COURSE (CNO, CNAME) values (2, ‘ORACLE开发’);
insert into T_COURSE (CNO, CNAME) values (3, ‘C++程序设计’);
insert into T_COURSE (CNO, CNAME) values (4, ‘C#程序设计’);
insert into T_COURSE (CNO, CNAME) values (5, ‘Windows实战’);
insert into T_COURSE (CNO, CNAME) values (6, ‘Center OS教程’);
insert into T_COURSE (CNO, CNAME) values (7, ‘Jsp/Servlet开发’);
insert into T_COURSE (CNO, CNAME) values (8, ‘J2EE从入门到精通’);
insert into T_COURSE (CNO, CNAME) values (9, ‘EJB及设计模式’);
insert into T_COURSE (CNO, CNAME) values (10, ‘Javascript/jQuery实战’);
insert into T_COURSE (CNO, CNAME) values (11, ‘Flash设计’);
insert into T_COURSE (CNO, CNAME) values (12, ‘HTML/CSS/JAVASCRIPT实战’);
insert into T_COURSE (CNO, CNAME) values (13, ‘精通ASP.NET’);
insert into T_COURSE (CNO, CNAME) values (14, ‘JBoss入门’);
insert into T_COURSE (CNO, CNAME) values (15, ‘Spring开发’);
commit;

–初始化成绩表
insert into T_SCORE (SNO, CNO, GRADE, TNO, ID) values (1, 2, 90.0, 2, 1);
insert into T_SCORE (SNO, CNO, GRADE, TNO, ID) values (1, 3, 80.0, 3, 2);
insert into T_SCORE (SNO, CNO, GRADE, TNO, ID) values (1, 4, 90.0, 4, 3);
insert into T_SCORE (SNO, CNO, GRADE, TNO, ID) values (11, 2, 70.0, 2, 4);
insert into T_SCORE (SNO, CNO, GRADE, TNO, ID) values (2, 11, 66.0, 11, 5);
insert into T_SCORE (SNO, CNO, GRADE, TNO, ID) values (2, 15, 77.0, 15, 6);
insert into T_SCORE (SNO, CNO, GRADE, TNO, ID) values (2, 8, 87.0, 8, 7);
insert into T_SCORE (SNO, CNO, GRADE, TNO, ID) values (2, 6, 96.0, 6, 8);
insert into T_SCORE (SNO, CNO, GRADE, TNO, ID) values (3, 2, 89.0, 2, 9);
insert into T_SCORE (SNO, CNO, GRADE, TNO, ID) values (3, 1, 91.0, 1, 10);
insert into T_SCORE (SNO, CNO, GRADE, TNO, ID) values (4, 2, 83.0, 2, 11);
insert into T_SCORE (SNO, CNO, GRADE, TNO, ID) values (5, 4, 73.0, 4, 12);
insert into T_SCORE (SNO, CNO, GRADE, TNO, ID) values (5, 1, 60.0, 1, 13);
insert into T_SCORE (SNO, CNO, GRADE, TNO, ID) values (5, 8, 82.0, 8, 14);
insert into T_SCORE (SNO, CNO, GRADE, TNO, ID) values (6, 8, 90.5, 10, 15);
insert into T_SCORE (SNO, CNO, GRADE, TNO, ID) values (8, 2, 58.0, 2, 16);
insert into T_SCORE (SNO, CNO, GRADE, TNO, ID) values (11, 3, 80.0, 3, 17);
insert into T_SCORE (SNO, CNO, GRADE, TNO, ID) values (9, 11, 65.0, 11, 18);
insert into T_SCORE (SNO, CNO, GRADE, TNO, ID) values (9, 12, 67.0, 12, 19);
insert into T_SCORE (SNO, CNO, GRADE, TNO, ID) values (9, 15, 95.0, 15, 20);
insert into T_SCORE (SNO, CNO, GRADE, TNO, ID) values (9, 13, 59.0, 13, 21);
insert into T_SCORE (SNO, CNO, GRADE, TNO, ID) values (10, 4, 98.0, 4, 22);
insert into T_SCORE (SNO, CNO, GRADE, TNO, ID) values (10, 6, 97.0, 6, 23);
insert into T_SCORE (SNO, CNO, GRADE, TNO, ID) values (10, 7, 96.0, 7, 24);
insert into T_SCORE (SNO, CNO, GRADE, TNO, ID) values (11, 7, 95.0, 7, 25);
insert into T_SCORE (SNO, CNO, GRADE, TNO, ID) values (12, 8, 69.0, 8, 26);
insert into T_SCORE (SNO, CNO, GRADE, TNO, ID) values (12, 9, 85.0, 9, 27);
insert into T_SCORE (SNO, CNO, GRADE, TNO, ID) values (10, 14, 100.0, 14, 28);
insert into T_SCORE (SNO, CNO, GRADE, TNO, ID) values (6, 9, 100.0, 9, 29);
insert into T_SCORE (SNO, CNO, GRADE, TNO, ID) values (11, 1, 59.0, 1, 30);
insert into T_SCORE (SNO, CNO, GRADE, TNO, ID) values (11, 4, 90.0, 4, 31);
insert into T_SCORE (SNO, CNO, GRADE, TNO, ID) values (11, 5, 91.0, 5, 32);
insert into T_SCORE (SNO, CNO, GRADE, TNO, ID) values (11, 6, 58.0, 6, 33);
insert into T_SCORE (SNO, CNO, GRADE, TNO, ID) values (11, 8, 93.0, 8, 34);
insert into T_SCORE (SNO, CNO, GRADE, TNO, ID) values (11, 9, 57.0, 9, 35);
insert into T_SCORE (SNO, CNO, GRADE, TNO, ID) values (11, 10, 95.0, 10, 36);
insert into T_SCORE (SNO, CNO, GRADE, TNO, ID) values (11, 11, 96.0, 11, 37);
insert into T_SCORE (SNO, CNO, GRADE, TNO, ID) values (11, 12, 97.0, 12, 38);
insert into T_SCORE (SNO, CNO, GRADE, TNO, ID) values (11, 13, 98.0, 13, 39);
insert into T_SCORE (SNO, CNO, GRADE, TNO, ID) values (11, 14, 99.0, 14, 40);
insert into T_SCORE (SNO, CNO, GRADE, TNO, ID) values (11, 15, 89.0, 15, 41);
commit;
在思维举重前,先来一个热身运动吧:

–(一)查询选修课程名称为’JAVA程序设计’的学员学号和姓名
–方法一:
–1.查询’JAVA程序设计’这门课程的课程号
–2.在成绩表中查询课程号为步骤1的课程号的成绩信息..
–3.在学生表中查询学号为步骤2结果的学生信息
select st.sno, st.sname
from student st
where st.sno in
(select sno
from score sc
where sc.cno =
(select co.cno from course co where co.cname = ‘JAVA程序设计’));

–方法二:
–1.查询’JAVA程序设计’这门课程的课程号
–2.在成绩表和学生表关联结果中查询课程号等于步骤1的课程号的成绩信息.
–sql86
select st.sno, st.sname
from student st, score sc
where st.sno = sc.sno
and sc.cno =
(select co.cno from course co where co.cname = ‘JAVA程序设计’);

–sql92
select st.sno, st.sname
from student st
join score sc
on st.sno = sc.sno
where sc.cno =
(select co.cno from course co where co.cname = ‘JAVA程序设计’);

–方法三:
–1.在成绩表和课程表关联结果中查询选修了’JAVA程序设计’这门课的学生的学号
–2.在学生表中查询步骤1中的学号的学生的详细信息.
–sql86
select st.sno, st.sname
from student st
where st.sno in (select sc.sno
from score sc, course co
where sc.cno = co.cno
and co.cname = ‘JAVA程序设计’);

–sql92
select st.sno, st.sname
from student st
where st.sno in (select sc.sno
from score sc
join course co
on sc.cno = co.cno
where co.cname = ‘JAVA程序设计’);

–方法四:
–在成绩表和课程表和学生表三表关联的结果中过滤得到选修了 ‘JAVA程序设计’的学生基本信息
–sql86
select st.sno, st.sname
from student st, score sc, course co
where st.sno = sc.sno
and co.cno = sc.cno
and co.cname = ‘JAVA程序设计’;

–sql92(1) 用where过滤
select st.sno, st.sname
from student st
join score sc
on st.sno = sc.sno
join course co
on co.cno = sc.cno
where co.cname = ‘JAVA程序设计’;

–sql92(2) 在关联条件中过滤
select st.sno, st.sname
from student st
join score sc
on st.sno = sc.sno
join course co
on co.cno = sc.cno
and co.cname = ‘JAVA程序设计’;

注: 1.对于sql86 和sql92的区别见这篇文章: http://blog.csdn.net/lk_blog/article/details/7580300
2.如果您在看的过程中觉得不熟悉的地方很多,建议您先看此文: http://blog.csdn.net/lk_blog/article/details/7585501

进入正文:

–(一)查询不选修课程编号为’1’的学员姓名和所属单位
–1.在成绩表中查询课程号为’1’的所有学生学号
–2.在学生表中查询学号不在步骤1中的学生的基本信息.
select st.sname, st.sdree
from t_student st
where st.sno not in (select sc.sno from t_score sc where sc.cno = ‘1’);

–(二)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 sc.sno, st.sname
from t_score sc
join t_student st
on sc.sno = st.sno
group by sc.sno, st.sname
having avg(sc.grade) > 85;

–(三)查询课程名称为”JAVA程序设计”,且分数低于60的学生姓名和分数
select *
from t_score sc
join t_course co
on sc.cno = co.cno
where sc.grade < 60
and co.cname = ‘JAVA程序设计’

我的代码:
select st.sname,sc.score from t_student stu,t_score sc,t_course co where st.stuno=sc.stuno and sc.score<60 and sc.name=’java程序设计’;

select *
from t_score sc
join t_course co
on sc.cno = co.cno
and sc.grade < 60
and co.cname = ‘JAVA程序设计’

–(四)查询任何一门课程成绩全部都在70分以上的姓名、课程名称和分数?
–1.查询出成绩小于70分的学生的学号.
–2.将学生,成绩,课程三张表作关联.
–3.在关联表中过滤出不在步骤1查询结果中的学生信息.
select st.sname, co.cname, sc.grade
from t_student st
join t_score sc
on st.sno = sc.sno
join t_course co
on sc.cno = co.cno
where st.sno not in (select sc1.cno from t_score sc1 where sc1.grade < 70);

–(五)查询出选了课的学生的人数.
select count(distinct(sc.sno)) from t_score sc;
–(六)查询每门课程被选课的学生数
select sc.cno, count(distinct(sc.sno)) from t_score sc group by sc.cno;
–(七)查询选了全部课程的学员姓名和所属单位
–1.在课程表中查询出所有课程的数量
–2.在成绩表中查询出学生选课数等于步骤1中总选课数的学生的学号,注意要用distinct,having中可以使用count,where中不能使用count.
–3.在学生表中查出步骤2中学生的基本信息.
select st.sname, st.sdree
from t_student st
where st.sno in
(select sc.sno
from t_score sc
group by sc.sno
having count(distinct sc.cno) = (select count(distinct sc1.cno)
from t_course sc1));

–(八) 查询没有学全所有课的同学的学号、姓名
select st.sname, st.sdree
from t_student st
where st.sno in
(select sc.sno
from t_score sc
group by sc.sno
having count(distinct sc.cno) != (select count(*) from t_course));

–(九)查询选修课程超过5门的学员学号和所属单位
–1.在成绩表中查询出选课程超过5门的学生学号.
–2.在学生表中查询步骤1中学号的学生的基本信息.
select st.sname, st.sdree
from t_student st
where st.sno in (select sc.sno
from t_score sc
group by sc.sno
having count(distinct sc.cno) > 5);

–(十)查询出没有选课的学生基本信息
–1.在成绩表中查询出所有选过课的学生的学号.
–2.在学生表中查询出步骤1中学生的基本信息.
select *
from t_student st
where st.sno not in (select sc.sno from t_score sc);

–下面的两个sql等价,在成绩表中数据量很大时使用下面的sql
select *
from t_student st
where st.sno not in (select distinct (sc.sno) from t_score sc);

select *
from t_student st
where st.sno not in (select sc.sno from t_score sc group by sc.sno);
–(十一) 列出有二门以上不及格课程的学生姓名及其平均成绩
–方法一
–1.在成绩表中查询出2门不及格学生的学号,结果记作t1
–2.将学生表和t1和成绩表三表作关联得到关联表,在关联表中取学生基本信息和平均成绩.
–sql92
select st.sno, st.sname, avg(sc.grade)
from t_student st
join (select sc.sno
from t_score sc
where sc.grade < 60
group by sc.sno
having count(distinct sc.cno) > 2) t1
on st.sno = t1.sno
join t_score sc
on sc.sno = t1.sno
group by st.sno, st.sname;

–sql86
select st.sno, st.sname, avg(sc.grade)
from t_student st,
t_score sc,
(select sc.sno
from t_score sc
where sc.grade < 60
group by sc.sno
having count(distinct sc.cno) > 2) t1
where st.sno = t1.sno
and sc.sno = t1.sno
group by st.sno, st.sname;

–方法二:
–1.在成绩表中查询出2门不及格学生的学号
–2.将学生表和成绩表通过学号作关联并根据步骤1中的结果作过滤,在关联结果中取出学生基本信息和平均成绩
select st.sno, st.sname, avg(sc.grade)
from t_student st
join t_score sc
on st.sno = sc.sno
where st.sno in (select sc.sno
from t_score sc
where sc.grade < 60
group by sc.sno
having count(distinct sc.cno) > 2)
group by st.sno, st.sname;
–(十二) 查询平均成绩大于60分的同学的学号和平均成绩
–学生表和课程表关联,在having子句中过滤平均成绩大于60分.
select st.sno, avg(sc.grade)
from t_student st, t_score sc
where st.sno = sc.sno
group by st.sno
having avg(sc.grade) > 60;

–1.学生表和课程表关联,将关联的结果记作t1
–2.在t1中过滤平均成绩大于60的学生学号.
select t1.sno, t1.avg_grade
from (select st.sno, avg(sc.grade) avg_grade
from t_student st, t_score sc
where st.sno = sc.sno
group by st.sno) t1
where t1.avg_grade > 60;

–(十三)查询出只选修了一门课程的全部学生的学号和姓名
–方法一:
–1.将学生表和成绩表作关联,在分组函数中使用having子句过滤出只选了一门课程的学生基本信息.
select sc.sno, st.sname
from t_score sc
join t_student st
on sc.sno = st.sno
group by sc.sno, st.sname
having count(distinct sc.cno) = 1;

–方法二:
–1.在成绩表中查找学号,分组函数的过滤条件判断只选择了一门课程的学生.
–2.在学生表中查找学号在步骤1中的值的学生的基本信息
select st.sno,st.sname
from t_student st
where st.sno in (select sc.sno
from t_score sc
group by sc.sno
having count(distinct sc.cno) = 1);

–(十四)查询至少有一门课与学号为”1″的同学所学相同的同学的学号和姓名
select st.sno, st.sname
from t_student st
join t_score sc1
on st.sno = sc1.sno
where sc1.cno in (select sc.cno from t_score sc where sc.sno = ‘1’)
group by st.sno, st.sname;

 

–(十五)列出既学过”1″号课程,又学过”2″号课程的所有学生姓名
–1.将成绩表和课程表作关联,在关联条件中作过滤查询出既选过课程’1’又选过课程’2’的学生的学号,注意看 co.cno in (‘1’, ‘2’)和having count(distinct sc.cno) = 2 的位置.
–2.在学生表中根据步骤1的结果作过滤查询出学生的基本信息.
–方法一:
–sql86
select st.sno, st.sname
from t_student st,
(select sc.sno
from t_score sc, t_course co
where sc.cno = co.cno
and co.cno in (‘1’, ‘2’)
group by sc.sno
having count(distinct sc.cno) = 2) t1
where st.sno = t1.sno;

–sql92
select st.sno, st.sname
from t_student st join
(select sc.sno
from t_score sc join t_course co
on sc.cno = co.cno
and co.cno in (‘1’, ‘2’)
group by sc.sno
having count(distinct sc.cno) = 2) t1
on st.sno = t1.sno;

–方法二:
–sql86
select st.sno, st.sname
from t_student st
where st.sno in (select sc.sno
from t_score sc, t_course co
where sc.cno = co.cno
and co.cno in (‘1’, ‘2’)
group by sc.sno
having count(distinct sc.cno) = 2);

–sql92
select st.sno, st.sname
from t_student st
where st.sno in (select sc.sno
from t_score sc
join t_course co
on sc.cno = co.cno
and co.cno in (‘1’, ‘2’)
group by sc.sno
having count(distinct sc.cno) = 2);

–(十六)查询至少学过学号为”1″的同学所有门课的同学学号和姓名
–1.查询出’1’号同学学习的全部课程.
–2.查询出’1’号同学学习全部课程的数量.
–3.将课程表和成绩表做关联,在关联表中查询出学生的学号,关联条件中加入过滤条件[课程号在步骤1查询结果范围内],过滤条件中加入数量等级步骤2中得到的数量.
–4.在学生表中查询步骤3中的学号的学生的基本信息.
select st.sno, st.sname
from t_student st
where st.sno in
(select sc.sno
from t_score sc
join t_course co
on sc.cno = co.cno
and co.cno in (select sc.cno from t_score sc where sc.sno = ‘1’)
group by sc.sno
having count(distinct sc.cno) = (select count(distinct sc.cno)
from t_score sc
where sc.sno = ‘1’))

–(十七)查询和”6″号同学学习的课程完全相同的同学的学号和姓名
–分析:要查询与6号同学完全相同的课程的学生信息,等价于学过6号同学的学过的所有课程并且选课数量与6同学选课数量相等.
–方法一:
–1.查询出’1’号同学学习的全部课程.
–2.查询出’1’号同学学习全部课程的数量.
–3.将课程表和成绩表做关联,在关联表中查询出学生的学号和选课数量,记作 t2,关联条件中加入过滤条件[课程号在步骤1查询结果范围内],过滤条件中加入数量等级步骤2中得到的数量.
–4.在成绩表中查询出学号和每个学生选课数量.得到结果记作: t1
–5.将步骤3中的t2和步骤4中的t1通过学生学号关联,添加过滤条件,t1中的选课数量等于t2中的选课数量.
–6.在学生表中查询不步骤5的学生学号的基本信息.
select st.sno, st.sname
from t_student st
where st.sno in
(select t1.sno
from (select sc_a.sno, count(distinct sc_a.cno) num_outer
from t_score sc_a
group by sc_a.sno) t1
join (select sc.sno, count(distinct sc.cno) num_inner
from t_score sc
join t_course co
on sc.cno = co.cno
and co.cno in
(select sc.cno from t_score sc where sc.sno = ‘6’)
group by sc.sno
having count(distinct sc.cno) = (select count(distinct sc.cno)
from t_score sc
where sc.sno = ‘6’)) t2
on t1.sno = t2.sno
where t1.num_outer = t2.num_inner);

–(十八)列出”1″号课成绩比”2″号课成绩高的所有学生的学号及其”1″号课和”2″号课的成绩
–1.将学生表和课程表作两次关联,一次关联用于取该学生课程’1’的成绩,另一次关联用于取该学生课程’2’的成绩.
–sql86
select st.sno, st.sname, sc_a.grade, sc_b.grade
from t_student st, t_score sc_a, t_score sc_b
where sc_a.cno = ‘1’
and sc_b.cno = ‘2’
and st.sno = sc_a.sno
and st.sno = sc_b.sno
and sc_a.grade > sc_b.grade

–sql92
select st.sno, st.sname, sc_a.grade, sc_b.grade
from t_student st
join t_score sc_a
on st.sno = sc_a.sno
join t_score sc_b
on st.sno = sc_b.sno
where sc_a.cno = ‘1’
and sc_b.cno = ‘2’
and sc_a.grade > sc_b.grade
–(十九)查询所有同学的学号、姓名、选课数、总成绩
select st.sno, st.sname, count(sc.cno), sum(sc.grade)
from t_student st, t_score sc
where st.sno = sc.sno
group by st.sno, st.sname;

 

–(二十)查询课程成绩小于60分的同学的学号,姓名,课程名,成绩
–下面两条sql虽然结果相同,但意义不同,注意理解一下哦.
–1.将学生表,课程表,成绩表作关联
–2.对关联后的结果作过滤,过滤出成绩小于60的学生基本信息.
select st.sno, st.sname, co.cname, sc.grade
from t_score sc
join t_student st
on sc.sno = st.sno
join t_course co
on sc.cno = co.cno
where sc.grade < 60

–1.将学生表,课程表,成绩表作关联,在关联条件中过滤成绩小于60.
select st.sno, st.sname, co.cname, sc.grade
from t_score sc
join t_student st
on sc.sno = st.sno
join t_course co
on sc.cno = co.cno
and sc.grade <60

 

–(二十一)按平均成绩从到低显示所有学生的”JAVA程序设计”、”J2EE从入门到精通”、”EJB及设计模式”三门的课程成绩,
–并按如下形式显示: 学生ID,姓名,JAVA程序设计,J2EE从入门到精通,EJB及设计模式,有效课程数,有效课程平均分
–1.将成绩表和课程表关联得到结果记作: t1, 关联时的条件选择只统计以上三门课程.
–2.按题目中的要求组织统计结果.
select st.sno,
st.sname,
sum(decode(t1.cname, ‘JAVA程序设计’, t1.grade)) JAVA程序设计,
sum(decode(t1.cname, ‘J2EE从入门到精通’, t1.grade)) J2EE从入门到精通,
sum(decode(t1.cname, ‘EJB及设计模式’, t1.grade)) EJB及设计模式,
count(distinct t1.grade) 有效课程数,
avg(t1.grade) 有效课程平均分
from t_student st
join (select *
from t_score sc
join t_course co
on sc.cno = co.cno
and co.cname in
(‘JAVA程序设计’, ‘J2EE从入门到精通’, ‘EJB及设计模式’)) t1
on st.sno = t1.sno
group by st.sno, st.sname

–将decode可以换成case when 第一种形式
select st.sno,
st.sname,
sum(case t1.cname
when ‘JAVA程序设计’ then
t1.grade
end) JAVA程序设计,
sum(case t1.cname
when ‘J2EE从入门到精通’ then
t1.grade
end) J2EE从入门到精通,
sum(case t1.cname
when ‘EJB及设计模式’ then
t1.grade
end) EJB及设计模式,
count(distinct t1.grade) 有效课程数,
avg(t1.grade) 有效课程平均分
from t_student st
join (select *
from t_score sc
join t_course co
on sc.cno = co.cno
and co.cname in
(‘JAVA程序设计’, ‘J2EE从入门到精通’, ‘EJB及设计模式’)) t1
on st.sno = t1.sno
group by st.sno, st.sname

–将decode可以换成case when 第二种形式
select st.sno,
st.sname,
sum(case
when t1.cname = ‘JAVA程序设计’ then
t1.grade
end) JAVA程序设计,
sum(case
when t1.cname = ‘J2EE从入门到精通’ then
t1.grade
end) J2EE从入门到精通,
sum(case
when t1.cname = ‘EJB及设计模式’ then
t1.grade
end) EJB及设计模式,
count(distinct t1.grade) 有效课程数,
avg(t1.grade) 有效课程平均分
from t_student st
join (select *
from t_score sc
join t_course co
on sc.cno = co.cno
and co.cname in
(‘JAVA程序设计’, ‘J2EE从入门到精通’, ‘EJB及设计模式’)) t1
on st.sno = t1.sno
group by st.sno, st.sname

–(二十二)查询各科成绩最高和最低的分:以如下形式显示:课程ID,课程名,最高分,最低分
select sc.cno, co.cname, max(grade), min(grade)
from t_score sc
join t_course co
on sc.cno = co.cno
group by sc.cno, co.cname
–(二十三)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
–1.在成绩表中查出课程号,平均成绩,课程人数,记作 : t1
–2.在成绩表中查出课程号,及格的课程人数,记作 : t2
–3.将步骤1中的成绩和步骤2中的课程2关联,查出所要的结果并排序.
select t1.cno, t1.avg_num 平均成绩, (count_num1 / count_num) * 100 及格率
from (select sc.cno, avg(grade) avg_num, count(distinct sc.sno) count_num
from t_score sc
group by sc.cno) t1
join (select sc1.cno, count(distinct sc1.sno) count_num1
from t_score sc1
where sc1.grade > 60
group by sc1.cno) t2
on t1.cno = t2.cno
order by t1.avg_num asc, 及格率 desc

–(二十四)统计各科成绩,各分数段人数:课程ID,课程名称,[100-90]优,[90-80]良,[80-70]中,[70-60]一般,[<60]不及格
–1.在成绩表中根据成绩值分段
–2.将步骤1中的结果与课程表关联.
select sc.cno,co.cname,
sum(case
when sc.grade > 90 then
1
end) 优,
sum(case
when sc.grade > 80 and sc.grade < 90 then
1
end) 良,
sum(case
when sc.grade > 70 and sc.grade < 80 then
1
end) 中,
sum(case
when sc.grade > 60 and sc.grade < 70 then
1
end) 一般,
sum(case
when sc.grade < 60 then
1
end) 不及格
from t_score sc join t_course co on sc.cno = co.cno
group by sc.cno,co.cname

–(二十五)查询学生平均成绩及其名次
select st.sno, st.sname, avg(sc.grade) avg_num
from t_score sc
join t_student st
on sc.sno = st.sno
group by st.sno, st.sname
order by avg_num desc

 

–(二十六)查询课程号分别为1,2,3的课程,成绩前三名的学生基本信息:(不考虑成绩并列情况)
–方法一:
–1.分别查出1,2,3各自的前3名的学生的学号,并用union all将结果集关联.
–2.在学生表中查询步骤1中查到的id的学生的基本信息.
select *
from t_student
where sno in (select t1.sno
from (select sc1.*
from t_score sc1
where sc1.cno = 1
order by sc1.grade desc) t1
where rownum < 4
union all
select t1.sno
from (select sc1.*
from t_score sc1
where sc1.cno = 2
order by sc1.grade desc) t1
where rownum < 4
union all
select t1.sno
from (select sc1.*
from t_score sc1
where sc1.cno = 3
order by sc1.grade desc) t1
where rownum < 4)

–方法二:
–rank() over(Partition .. order by …) 是按照某个字段的值进行分组并编号
select t1.cno, t1.sno, t1.grade, r
from (select sc.sno,
sc.cno,
sc.grade,
rank() over(partition by sc.cno order by grade desc) r
from t_score sc) t1
where r < 4
and t1.cno in (1, 2, 3)
order by t1.cno, t1.sno, r;
–(二十七)查询各科成绩前三名的记录(不考虑成绩并列情况)
–rank() over(Partition .. order by …) 是按照某个字段的值进行分组并编号
select t1.cno, t1.sno, t1.grade, r
from (select sc.sno,
sc.cno,
sc.grade,
rank() over(partition by sc.cno order by grade desc) r
from t_score sc) t1
where r < 4
order by t1.cno, t1.sno, r;

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