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
