一．实验目的：
熟练掌握使用SELECT语句进行数据查询。
二．实验内容：（所有题写到实验报告中）
1.对数据库stuinfo使用T-SQL命令进行如下操作：
查询student表中的学号、姓名和年龄并为列设置别名，结果按学号升序排。查询班级（要求不重复）。查询姓“王”的学生信息。查询成绩在80-100之间的学号、课程号及成绩，结果先按课程号升序排，课程号一样的再按成绩降序排。查询所有缺考的学生的学号及课程号。查询 ‘3-105’课的选课人数、最高分、最低分和平均分。查询每位同学的平均成绩，包括学号和平均成绩两列，结果按学号升序排。查询各班各门课的平均成绩，包括班号、课程号和平均成绩三列，结果先按班升序排，班一样的再按课程号升序排。查询score表中至少有5名学生选修的课程号及平均分。查询其平均分高于80的学生的学号，姓名和平均分。
查询“95031”班所选课程的平均分，包括课程名和平均分。
查询所有教师的任课情况，包括教师姓名和课程名两列，如果某位教师没有任课则课程名列显示NULL。
查询其最低分大于70，最高分小于90的学生的学号、所选课程号及其分数。
查询成绩高于所选课平均分的学生学号、姓名、课程号和成绩。
查询每门课最高分的课程名、学生姓名和成绩。
查询选修其课程的学生人数多于5人的教师姓名。
查询没选“张旭”教师课的学生成绩，并按成绩递增排列。
查询没有任课的教师的姓名。
查询没有选修”6-166″课程的学生的学号和姓名。
查询出所有男生信息放入NS表中。
删除没人选的课程。
将“95031”班学生的成绩全部减去10分。
\2. 对订单管理库ordermanagement使用T-SQL命令进行下列查询。
ordermanagement数据库中有三个表，其结构如下：（加下划线的为主键）
客户表customer(客户号，客户名，地址，电话)
订单表order_list(订单号，客户号，订购日期)
订单明细表Order_detail(订单号，器件号，器件名，单价，数量)
使用SELECT语句完成下列查询：
查询2001年的所有订单信息（订单号，客户号，订购日期）。
查询订单明细中有哪些器件（即查询不重复的器件号和器件名）。
查询客户名为“三益贸易公司”的订购单明细（订单号、器件号、器件名、单价和数量），
查询结果先按“订单号”升序排，同一订单的再按“单价”降序排。
查询目前没有订单的客户信息。
查询客户名中有“科技”字样的客户信息。
查询每笔订单的订单号和总金额，查询结果按订单号升序排，查询结果存入表ZJE中。
查询订单数量超过5笔的客户号及订单数量，查询结果按订单数量降序排。
查询每种器件中单价最低的订单明细（订单号、器件号、器件名、单价和数量）。
对表order_detail建立查询，把“订单号”的尾部字母相同且“器件号”相同的订单合并
成一张订单，新的“订单号”取原来订单号的尾部字母，器件号不变，“单价”取最低价，
“数量”取合计，查询结果先按新的“订单号”升序排，再按“器件号”升序排。
- 查询销售总数量最多的三种器件及其总数量。
use stuinfo
select sno as ‘学号’,sname as ‘姓名’,DateName(year,GetDate())-DATEPART(YYYY,sbirthday) as ‘年龄’ from student order by sno asc;
select sno as ‘学号’,sname as ‘姓名’, DateName(year,GetDate())-DATEPART(YYYY,sbirthday) as ‘年龄’ from student order by sno desc
—2
select distinct sclass from student
—3
select sname from student where substring(sname,1,1)=’王’
—4
select sno,cno,degree from score where degree>=80 order by cno asc,degree desc
—5
select sno,cno from score where degree is null
—6
select COUNT(*) as ‘选课人数’,MAX(degree) as ‘max’,MIN(degree) as ‘min’,AVG(degree) as ‘avg’ from score
—7
select sno,AVG(degree)as ‘avg’ from score group by sno order by sno desc
—8
select sclass,cno,AVG(degree) as ‘AVG’ from student,score group by sclass,cno order by sclass,cno
—9
select cno,AVG(degree) from score group by cno having COUNT(*) >= 5
—10
select student.sno as ‘学号’, sname as ‘姓名’ ,AVG(degree) as ‘AVG’ from student,score where student.sno=score.sno group by student.sno,student.sname having AVG(degree)>=80
—11
select course.cname as ‘课程名’,AVG(degree) as ‘平均分’ from student,course,score where student.sno=score.sno and sclass=’95031′ group by course.cno,course.cname
—12
select tname as ‘教师姓名’,cname as ‘课程名’ from teacher left join course on teacher.tno = course.tno group by tname,cname
—13
select sno as ‘学号’,cno as ‘课程号’,degree as ‘分数’ from score a WHERE (SELECT MIN(degree) FROM score b where a.sno=b.sno)>70 and (SELECT MAX(degree)FROM score c where a.sno=c.sno)<90 and degree is not NUll
—14
select score.sno,sname,cno,degree from student,score where student.sno=score.sno and degree>(select AVG(degree) from score,course where score.cno=course.cno)
—15
select cname ‘课程名’,sname ‘姓名’,degree ‘成绩’ from student,course,score where student.sno=score.sno and course.cno=score.cno and degree=(select MAX(degree)from score where score.cno=course.cno)
—16
select tname as 教师姓名 from teacher,course where teacher.tno=course.tno and (select COUNT(*) from score where score.cno=course.cno)>5
—17
select score.degree from score,teacher,course where teacher.tno=course.tno and course.cno=score.cno and tname!=’张旭’ order by degree
—18
select tname from teacher except select tname from teacher,course where teacher.tno=course.tno
—19
select student.sno,student.sname from student,course,score where score.sno not in (select sno from score where cno=’1-166′) group by student.sno,student.sname
—20
select student.sno as 学号,student.sname as 学生姓名,ssex as 性别,sbirthday as 生日,sclass as 班级,cname as 课程,degree as 成绩 into NS from student,score,course where student.sno=score.sno and ssex=’男’ and course.cno=score.cno order by student.sno
—21
delete course where course.cno not in (select cno from score)
—22
update score set degree = degree – 10 from student,score where student.sno=score.sno and student.sclass=’95031′
第二题
use OrderManagement
—1
select * from order_list
—2
select distinct 器件号, 器件名 from order_detail
—3
select order_detail.订单号, order_detail.器件号, order_detail.器件名, order_detail.单价, order_detail.数量 from Customer, order_detail, order_list
where Customer.客户名 = ‘三益贸易公司’ and Customer.客户号 = order_list.客户号 and order_list.订单号 = order_detail.订单号
order by 订单号, 单价 desc
—4
select * from Customer where Customer.客户号 not in (select 客户号 from order_list)
—5
select * from Customer where 客户名 like ‘%科技%’
—6
select 订单号, sum(单价 * 数量) as ‘总金额’ into ZJE from order_detail group by 订单号 order by 订单号
—7
select 客户号, COUNT(订单号) from order_list group by 客户号 having count(订单号) > 5 order by COUNT(订单号) desc
—8
select * from order_detail a where 单价 in (select MIN(单价) from order_detail b where a.器件名=b.器件名)
—9
select RIGHT(订单号, 1) as 订单号, 器件号, MIN(单价)as 最低价,SUM(数量) as 数量 from order_detail group by 订单号, 器件号 order by 订单号, 器件号
—10
select top 3 器件号, 器件名, SUM(数量) as 数量 from order_detail group by 器件号, 器件名 order by 数量 desc
