select a.teaID,a.age count(*) from teacher a,student b,teacher_student c where a.teaID=c.teaID and b.stuID=c.stuID and a.age>40 and b.age>12 group by a.teaID,a.age;
select a.department,count from tA a,tB b where a.id=b.id group by b.id,a,deparment
方法1：
select Sname,sum(Ccredit) as totalCredit from Student,Course,SC where Grade>=60 and Student.Sno=SC.Sno and Course.Cno=SC.Cno group by Sname
select sname,sum(case when sc.grade<60 then 0 else course.Ccredit end) as totalCredit from Student,sc,course where sc.sno=student.sno and sc.cno=course.cno group by sname
select Sname,SUM(case when Grade<60 then 0 else Ccredit end) as totalGrade FROM SC JOIN Student ON(Student.sno = SC.sno) JOIN Course ON(SC.Cno = Course.Cno) GROUP BY Student.Sname;
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有3个表S，C，SC
S（SNO，SNAME）代表（学号，姓名）
C（CNO，CNAME，CTEACHER）代表（课号，课名，教师）
SC（SNO，CNO，SCGRADE）代表（学号，课号成绩）
问题：
1，找出没选过“黎明”老师的所有学生姓名。
2，列出2门以上（含2门）不及格学生姓名及平均成绩。
3，即学过1号课程又学过2号课所有学生的姓名。
请用标准SQL语言写出答案，方言也行（请说明是使用什么方言）。
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答案:
S（SNO，SNAME）代表（学号，姓名）
C（CNO，CNAME，CTEACHER）代表（课号，课名，教师）
SC（SNO，CNO，SCGRADE）代表（学号，课号成绩）
select sno,sname from s; select cno,cname,cteacher from c; select sno,cno,scgrade from sc;
问题1.找出没选过“黎明”老师的所有学生姓名。
第一步:求黎明老师教的所有课的课号
select distinct cno from c where cteacher='黎明'
第二步:选了黎明老师的所有学生的编号
select sno from sc where cno in (
)
第三步:没有选黎明老师的所有学生的姓名
select sname from s where sno not in (
)
即:
select sname from s where sno not in ( select sno from sc where cno in ( select distinct cno from c where cteacher='黎明' ) )
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问题2:列出2门以上（含2门）不及格学生姓名及平均成绩。
第一步:2门以上不及格的学生的学号
select sno from sc where scgrade < 60 group by sno having count(*) >= 2
第二步:每个学生平均分
select sno, avg(scgrade) as avg_grade from sc group by sno
第三步:第一步中得到的学号对应的学生姓名以及平均分
select s.sname ,avg_grade from s
即:
select s.sname ,avg_grade from s join (select sno, count(*) from sc where scgrade < 60 group by sno having count(*) >= 2)t on s.sno = t.sno join (select sno, avg(scgrade) as avg_grade from sc group by sno )t1 on s.sno = t1.sno
错误的写法:
错误在于：求的是所有不及格的课程的平均分，而不是所有课程（包括及格的）的平均分
执行顺序:
select sname, avg_scgrade from s join (select sno, avg(scgrade) avg_scgrade from sc where scgrade < 60 group by sno having count(*) >= 2) t on (s.sno = t.sno); ---------------------------------------------------------------------------- select sno,sname from s; select cno,cname,cteacher from c; select sno,cno,scgrade from sc;
问题3:即学过1号课程又学过2号课所有学生的姓名。
第一步:学过1号课程的学号
select sno from sc where cno = 1
第二步:学过2号课程的学号
select sno from sc where cno = 2
第三步:即学过1号课程又学过2号课的学号
select sno from sc where cno =1 and sno in (select sno from sc where cno = 2)
第四步:得到姓名
select sname from s where sno in ( select sno from sc where cno = 1 and sno in (select sno from sc where cno = 2) )
或者:
select sname from s where sno in (select sno from sc where cno = 1) and sno in (select sno from sc where cno = 2) company 公司名(companyname) 编号(id)
employeehired
2.找出表之间关系: 外键关系， employeehired (id) 参考 company (id)
3.求第四财季招聘过员工的公司名称:
select companyname from company c join employeehired e on (c.id = e.id) where fiscalquarter = 4;
select companyname from company where id not in (select distinct id from employeehired where fiscalquarter not in(1,2,3) );
select companyname , sum_numhired from company c join ( select sum(numhired) sum_numhired from employeehired group by id ) t on (c.sum_numhired = t.sum_numhired);
--求部门中哪些人的薪水最高----此处开始使用的是scott账户下的自带表
select ename, sal from emp join (select max(sal) max_sal, deptno from emp group by deptno) t on (emp.sal = t.max_sal and emp.deptno = t.deptno);
select deptno, avg_sal, grade from //从下面表中取，下表必须有字段 (select deptno, avg(sal) avg_sal from emp group by deptno) t join salgrade s on (t.avg_sal between s.losal and s.hisal);
--求每个部门的平均的薪水等级
select deptno, avg(grade) from (select deptno, ename, grade from emp join salgrade s on (emp.sal between s.losal and s.hisal)) t group by deptno;
--求雇员中有哪些人是经理人
select ename from emp where empno in (select distinct mgr from emp );
select distinct sal from emp where sal not in (select distinct e1.sal from emp e1 join emp e2 on (e1.sal < e2.sal));
--求平均薪水最高的部门的部门编号
select deptno, avg_sal from (select deptno, avg(sal) avg_sal from emp group by deptno) where avg_sal = (select max(avg_sal) from (select avg(sal) avg_sal, deptno from emp group by deptno) );
///////////另解../////////////////////////////
select deptno, avg_sal from (select deptno, avg(sal) avg_sal from emp group by deptno) where avg_sal = (select max(avg(sal)) from emp group by deptno);
////////组函数嵌套，不过只能套2层，因为多行输入，单行输出//////////
--求平均薪水最高的部门的部门名称
select dname from dept where deptno = ( select deptno from (select deptno, avg(sal) avg_sal from emp group by deptno) where avg_sal = (select max(avg_sal) from (select avg(sal) avg_sal, deptno from emp group by deptno) ) );
//从里到外
---先求出每个员工的薪水等级，然后再按照部门求出平均薪水等级
select avg_grade,deptno from (select avg(grade) avg_grade,deptno ( select grade,empno,deptno from emp e join salgrade s on(e.sal between s.losal adn s.hisal) ) group by deptno )
----完整的----
select empname ,avg_grade dept d join (select deptno,avg(grade) as avg_grade from (select deptno,empno,grade from emp e join salgrade s on(e.sal between s.losal and s.hisal) ) group by deptno )t1 on d.depno=t1.deptno;
1.平均薪水：select deptno, avg(sal) from emp group by deptno;
2.平均薪水的等级：
select deptno, grade, avg_sal from (select deptno, avg(sal) avg_sal from emp group by deptno) t join salgrade s on ( t.avg_sal between s.losal and s.hisal);
3.平均薪水最低的等级：
select min (grade) from ( select deptno, grade, avg_sal from (select deptno, avg(sal) avg_sal from emp group by deptno) t join salgrade s on ( t.avg_sal between s.losal and s.hisal) );
select dname, t1.deptno, grade, avg_sal from // deptno 未明确定义列 (select deptno, grade, avg_sal from (select deptno, avg(sal) avg_sal from emp group by deptno) t join salgrade s on ( t.avg_sal between s.losal and s.hisal) ) t1 join dept on (t1.deptno = dept.deptno) where t1.grade = ( select min (grade) from ( select deptno, grade, avg_sal from (select deptno, avg(sal) avg_sal from emp group by deptno) t join salgrade s on ( t.avg_sal between s.losal and s.hisal) ) ); //有完全重复的地方
：：：：：：：创建视图，视图就是表，子查询:虚表 ，链接：：：：：：：：
create view v$_dept_avg_sal_info as select deptno, grade, avg_sal from (select deptno, avg(sal) avg_sal from emp group by deptno) t join salgrade s on ( t.avg_sal between s.losal and s.hisal);
//视图已创建；
conn sys/10023 as sysdba;
grant create table, create view to sctt;
/////////默认是可以建表的；
select * from v$_dept_avg_sal_info;
5.化简
select dname, t1.deptno, grade, avg_sal from v$_dept_avg_sal_info t1 join dept on (t1.deptno = dept.deptno) where t1.grade = ( select min (grade) from v$_dept_avg_sal_info );
--求部门经理人中平均薪水最低的部门名称 (思考题)
--求比普通员工的最高薪水还要高的经理人名称
1.select distinct mgr from emp; //king'mgr is null; 2.select max(sal) from emp where empno not in (select distinct mgr from emp where mgr is not null); 3.select ename from emp where empno in (select distinct mgr from emp where mgr is not null) and sal > ( select max(sal) from emp where empno not in (select distinct mgr from emp where mgr is not null) );
--求薪水最高的前5名雇员
--求薪水最高的第6到第10名雇员(重点掌握)
--练习： 求最后入职的5名员工
--面试题： 比较效率
select * from emp where deptno = 10 and ename like '%A%'; select * from emp where ename like '%A%' and deptno = 10;
////////数字不对，后面就不用看了 ，先比较数字快；//也许Oracle有优化
//CSDN - 专家门诊 MS-SQL Server
:::::::::::::::::::::::::::::::::::回家作业:::::::::::::::::::::::::::::::::::::::::::
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1.按产品分类，仅列出各类商品中红色多于蓝色的商品名称及差额数量：
2.按产品分类，将数据按下列方式进行统计显示
来源：博客园
作者：java编程记录
链接：https://www.cnblogs.com/wuliaojava/p/11788283.html