SQL Server示例查询
Student(S#,Sname,Sage,Ssex) 学生表
Course(C#,Cname,T#) 课程表
SC(S#,C#,score) 成绩表
Teacher(T#,Tname) 教师表
问题:
1、查询“001”课程比“002”课程成绩高的所有学生的学号;
select a.S# from (select s#,score from SC where C#='001') a,(select s#,score
from SC where C#='002') b
where a.score>b.score and a.s#=b.s#;
2、查询平均成绩大于60分的同学的学号和平均成绩;
select S#,avg(score)
from sc
group by S# having avg(score) >60;
3、查询所有同学的学号、姓名、选课数、总成绩;
select Student.S#,Student.Sname,count(SC.C#),sum(score)
from Student left Outer join SC on Student.S#=SC.S#
group by Student.S#,Sname
4、查询姓“李”的老师的个数;
select count(distinct(Tname))
from Teacher
where Tname like '李%';
5、查询没学过“叶平”老师课的同学的学号、姓名;
select Student.S#,Student.Sname
from Student
where S# not in (select distinct( SC.S#) from SC,Course,Teacher where SC.C#=Course.C# and Teacher.T#=Course.T# and Teacher.Tname='叶平');
6、查询学过“001”并且也学过编号“002”课程的同学的学号、姓名;
select Student.S#,Student.Sname from Student,SC where Student.S#=SC.S# and SC.C#='001'and exists( Select * from SC as SC_2 where SC_2.S#=SC.S# and SC_2.C#='002');
7、查询学过“叶平”老师所教的所有课的同学的学号、姓名;
select S#,Sname
from Student
where S# in (select S# from SC ,Course ,Teacher where SC.C#=Course.C# and Teacher.T#=Course.T# and Teacher.Tname='叶平' group by S# having count(SC.C#)=(select count(C#) from Course,Teacher where Teacher.T#=Course.T# and Tname='叶平'));
8、查询课程编号“002”的成绩比课程编号“001”课程低的所有同学的学号、姓名;
Select S#,Sname from (select Student.S#,Student.Snam
相关文档:
xtype 代表类型
C = CHECK 约束
D = 默认值或 DEFAULT 约束
F = FOREIGN KEY 约束
L = 日志
FN = 标量函数
IF = 内嵌表函数
P = 存储过程
PK = PRIMARY KEY 约束(类型是 K)
RF = 复制筛选存储过程
S = 系统表
TF = 表函数
TR = 触发器
U = 用户表
UQ = UNIQUE 约束(类型是 K)
V = 视图 ......
行列倒置在sql server中是一种很常见的技巧,在做应用系统的时候,经常需要做一些统计功能避免不了使用行列倒置这个技巧,我小小的做了一下总结:
第一种:sql server 2000中使用case进行行列倒置
create table RowCellConvertTest
(
grade varchar(50),
sex varchar(50),
studentCount int
)
......
导出表结构:
Tools-->Export User Objects -->选择要导出的表(包括Sequence等)-->.sql文件
导出表数据:
Tools-->Export Tables-->选择表,选择SQL Inserts-->.sql文件
执行这些.sql文件时,要使用新建Command Window来执行. ......
select ks.login_name,ks.exam_name,ks.start_time,ks.end_time,cj.score
from (
select u.user_id,u.login_name,e.* from cphrms.EXAM_USER eu, cphrms.users u, cphrms.exam_info e
where eu.user_id = u.user_id and eu.exam_id = e.exam_id
) ks
left ......