group by 分组--每个工作的人数 select count(job) from emp group by 
group by 分组
--每个工作的人数
select count(job) from emp group by job;
--有多少工作
select count(coutn(job)) from emp group by job;
--每个工作的最高工资
select max(sal) from emp group by job;
--每个工作的最低工资
select min(sal) from emp group by job;
--每个工作的平均工资
select avg(sal) from emp group by job;
--所有工作中最大的平均工资
select max(avg(sal)) from emp group by job;
使用having 根据条件过滤组
select job,count(job) from emp group by job having count(job)>=3;
where 用在group by 之前
select job,count(job) from emp where sal>1000 group by job having count(job)>=3;
decode 函数
decode(value1 if1 then1 if2 then2 ... else)
decode(条件,值1,返回值1,值2,返回值2,...值n,返回值n,缺省值)
如果 value1=if1 返回then1
否则 如果 value1=if2 返回 then2
默认的返回值是else
select decode(sal,1600,'一千六',5000,'五千','其他') from emp;
--按照job的内容分类统计其中部门ID是30的人数
select job,count(decode(DEPTNO,30,1,null)) deptno from emp group by(job);
--按照部门分类 统计每个部门中工作是CLERK的人数
select deptno,count(decode(job,'CLERK',1,null)) counts from emp group by(deptno);
--按照部门分类 计算每个每部中工作是CLERK的平均工资
select deptno,avg(decode(job,'CLERK',sal,null)) CLERK_AVG_SAL from emp group by(deptno);
insert into emp values(8000,'lmdtx','CLERK',8080,to_date('1980/09/09','yyyy/mm/dd'),30000,null,null);
insert into emp values(8880,'dsy','CLERK',8980,to_date('1982/02/02','yyyy/mm/dd'),3000,null,null);
表的关联
外键 在一个表中 存着对应另外一张表的主键
连接表
1·笛卡尔积 emp 表中 deptno 为空的查询不到
--emp表中的所有的行* dept中的所有的行
select * from emp,dept;
--需要写过滤条件 过滤掉不需要的
select * from emp,dept where emp.deptno=dept.deptno;
select * from emp e.dept d where e.deptno=d.deptno;
2·内连接 这个结果和笛卡尔积 用条件后的结果一样
inner join 和 join 一样(省略 inner) 默认就是 内连接
join...on...
from emp e inner join dept d on e.deptno=d.deptno
两个表的一个外键的名字 和 另一张表的主键的名字一样
from emp inner join dept using(deptno)
select * from emp e inner join dept d on e.deptno=d.deptno;
select * from emp inner join dept using(deptno);
3·外连接
1·左外连(左边表的数据必须是全都有的,没有的地方补null)
from emp left outer join dept using(deptno)
select * from emp left outer join dept on emp.deptno=dept.deptno;
select ename,nvl(dname,'还没有部门') from emp2 left outer join dept2 using(deptno);
2·右外连
select * from emp right outer join dept using(deptno);
--查询所有部门的名称和部门人数按人数的降序排
select dname,count(empno) num
from emp2 right join dept2 Using(deptno)
group by dname
order by num desc;
3·全外连(全都有)
emp full outer join dept
select * from emp full outer join dept using(deptno);
--查询所有部门的名称以及部门的人数 没有部门的人 化为 未分配
select nvl(dname,'未分配'),count(empno)
from emp2 full outer join dept2 using(deptno)
group by dname
order by dname;
select nvl(dname,'未分配部门') 部门名字,decode(count(empno),0,'未分配人手',count(empno)) 人数
from emp2 full outer join dept2 using(deptno)
group by dname
order by dname;
4·自连接
select e1.ename,e2.ename from emp2 e1 inner join emp2 e2 on e1.mgr=e2.empno;
select e1.ename, e2.ename from emp2 e1 left join emp2 e2 on e1.mgr = e2.empno;
select nvl(e1.ename,'没有手下') , e2.ename MGR from emp2 e1 right join emp2 e2 on e1.mgr=e2.empno;
select e2.ename,decode(count(e1.ename),0,'没有手下',count(e1.ename))from emp2 e1 right join emp2 e2 on e1.mgr=e2.empno group by e2.ename
子查询
用在where 子句中,单行单列
select e1.ename from emp2 e1 inner join emp2 e2 on e1.mgr=e2.empno where e2.ename='KING';
select distinct dname from emp2 e1 join dept2 d2 ON e1.deptno= d2.deptno where e1.ename like '%S%';
select dname from dept2 where deptno in(select deptno from emp2 where ename like '%S%');
ANY() 任意一个
ALL() 所有
select ename,job,sal from emp2 where sal>all(select DISTINCT sal from emp2 where job='MANAGER') and job<>'MANAGER';
select ename job,sal from emp2 where sal>any(select distinct sal from emp2 where job='CLERK') and job<>'CLERK';
exists 用于判断后面的子查询是否有行 有true 没有false
select ename from emp2 e1 where EXISTS(select 1 from emp2 e2 where e2.mgr= e1.empno);
select ename from emp2 e1 where not exists(select 1 from emp2 e2 where e2.mgr=e1.empno);
3 in (1,2,3) true
3 in (1,2,null) false
5 not in (1,2,3) true
5 not in (1,2,null) false
5 in (1,2,null) false
5 not in (null) false
5in (null) false
not in 的子查询中不能出现null值
where 中的多列 查询
select e.ename,e.sal, d.deptno,d.dname from emp2 e inner join dept2 d on e.deptno= d.deptno where (e.deptno,e.sal) in (select deptno,min(sal) from emp2 where deptno is not null group by deptno);
select * from emp2 where (deptno,sal)in (select deptno,min(sal) from emp2 where deptno is not null group by deptno);
select deptno,min(sal) from emp2 where deptno is not null group by deptno;
用在having
from emp where deptno is not null group by deptno having avg(sal)<(select max(avg(sal)) from emp group by deptno)
select max(avg(sal)) from emp2 group by deptno;
select deptno,round(avg(sal)) from emp2 where deptno is not null group by deptno HAVING avg(sal)<(select max(avg(sal)) from emp2 group by deptno);
select dname,deptno from emp2 join dept2 using(deptno) where deptno is not null group by deptno,dname having avg(sal)=(select max(avg(sal)) from emp2 group by deptno);
分页
伪列
select empno,ename,rownum from emp2 order by ename;
select * from emp2 where rownum<=10;
select * from (select empno,ename, rownum n from emp2 where rownum<=10) where n>5 and n<=10
select * from (select empno,ename, rownum n from emp2 ) where n>5 and n<=10
select * from (select empno,ename, rownum n from emp2 where rownum<=10) where n>5
select * from (select empno,ename,sal, rownum n from (select * from emp2 order by sal) where rownum <=10) where n>5;
内联视图(Inner View)
就是把select后的结果当作一张表
select ename 名字,job 职位,emp_num 手下人数 from emp2 join(select deptno d,count(empno) emp_num from emp2 inner join dept2 using(deptno) group by deptno) on deptno=d where job='MANAGER';
select ename 名字,job 职位,dname 部门,deptno 部门编号,emp_num 手下人数 from emp2 inner join(select deptno d,count(empno) emp_num,dname from emp2 inner join dept2 using(deptno) group by deptno, dname) on deptno=d where job='MANAGER'
group by 对多个列分组
而这两个列都是对部门的唯一标识,不会影响分组
--结束END--
本文标题: Java38: 数据库二(Oracle)
本文链接: https://lsjlt.com/news/50939.html(转载时请注明来源链接)
有问题或投稿请发送至: 邮箱/279061341@qq.com QQ/279061341
2024-10-23
2024-10-22
2024-10-22
2024-10-22
2024-10-22
2024-10-22
2024-10-22
2024-10-22
2024-10-22
2024-10-22
回答
回答
回答
回答
回答
回答
回答
回答
回答
回答
0