oracle——笔记——分组查询 ——子查询

create table emp2 ( empno number(4),
                    ename varchar2(10),
                    job varchar2(9),
                    mgr number(4),
                    hiredate date,
                    sal number(7,2),
                    comm number(7,2),
                    deptno number(2) 
                 );
-----------------------------------------------------------------------------------------------------------

insert into emp2 (EMPNO, ENAME, JOB, MGR, HIREDATE, SAL, COMM, DEPTNO)
values (7369, ‘SMITH‘, ‘CLERK‘, 7902, to_date(‘17-12-1980‘, ‘dd-mm-yyyy‘), 800.00, null, 20);

insert into emp2 (EMPNO, ENAME, JOB, MGR, HIREDATE, SAL, COMM, DEPTNO)
values (7499, ‘ALLEN‘, ‘SALESMAN‘, 7698, to_date(‘20-02-1981‘, ‘dd-mm-yyyy‘), 1600.00, 300.00, 30);

insert into emp2 (EMPNO, ENAME, JOB, MGR, HIREDATE, SAL, COMM, DEPTNO)
values (7521, ‘WARD‘, ‘SALESMAN‘, 7698, to_date(‘22-02-1981‘, ‘dd-mm-yyyy‘), 1250.00, 500.00, 30);

insert into emp2 (EMPNO, ENAME, JOB, MGR, HIREDATE, SAL, COMM, DEPTNO)
values (7566, ‘JONES‘, ‘MANAGER‘, 7839, to_date(‘02-04-1981‘, ‘dd-mm-yyyy‘), 2975.00, null, 20);

insert into emp2 (EMPNO, ENAME, JOB, MGR, HIREDATE, SAL, COMM, DEPTNO)
values (7654, ‘MARTIN‘, ‘SALESMAN‘, 7698, to_date(‘28-09-1981‘, ‘dd-mm-yyyy‘), 1250.00, 1400.00, 30);

insert into emp2 (EMPNO, ENAME, JOB, MGR, HIREDATE, SAL, COMM, DEPTNO)
values (7698, ‘BLAKE‘, ‘MANAGER‘, 7839, to_date(‘01-05-1981‘, ‘dd-mm-yyyy‘), 2850.00, null, 30);

insert into emp2 (EMPNO, ENAME, JOB, MGR, HIREDATE, SAL, COMM, DEPTNO)
values (7782, ‘CLARK‘, ‘MANAGER‘, 7839, to_date(‘09-06-1981‘, ‘dd-mm-yyyy‘), 2450.00, null, 10);

insert into emp2 (EMPNO, ENAME, JOB, MGR, HIREDATE, SAL, COMM, DEPTNO)
values (7788, ‘SCOTT‘, ‘ANALYST‘, 7566, to_date(‘19-04-1987‘, ‘dd-mm-yyyy‘), 3000.00, null, 20);

insert into emp2 (EMPNO, ENAME, JOB, MGR, HIREDATE, SAL, COMM, DEPTNO)
values (7839, ‘KING‘, ‘PRESIDENT‘, null, to_date(‘17-11-1981‘, ‘dd-mm-yyyy‘), 5000.00, null, 10);

insert into emp2 (EMPNO, ENAME, JOB, MGR, HIREDATE, SAL, COMM, DEPTNO)
values (7844, ‘TURNER‘, ‘SALESMAN‘, 7698, to_date(‘08-09-1981‘, ‘dd-mm-yyyy‘), 1500.00, 0.00, 30);

insert into emp2 (EMPNO, ENAME, JOB, MGR, HIREDATE, SAL, COMM, DEPTNO)
values (7876, ‘ADAMS‘, ‘CLERK‘, 7788, to_date(‘23-05-1987‘, ‘dd-mm-yyyy‘), 1100.00, null, 20);

insert into emp2 (EMPNO, ENAME, JOB, MGR, HIREDATE, SAL, COMM, DEPTNO)
values (7900, ‘JAMES‘, ‘CLERK‘, 7698, to_date(‘03-12-1981‘, ‘dd-mm-yyyy‘), 950.00, null, 30);

insert into emp2 (EMPNO, ENAME, JOB, MGR, HIREDATE, SAL, COMM, DEPTNO)
values (7902, ‘FORD‘, ‘ANALYST‘, 7566, to_date(‘03-12-1981‘, ‘dd-mm-yyyy‘), 3000.00, null, 20);

insert into emp2 (EMPNO, ENAME, JOB, MGR, HIREDATE, SAL, COMM, DEPTNO)
values (7934, ‘MILLER‘, ‘CLERK‘, 7782, to_date(‘23-01-1982‘, ‘dd-mm-yyyy‘), 1300.00, null, 10);

=============================================================================================================================

create table dept2 ( deptno number(2),
                     dname varchar2(14),
                     loc varchar2(13)
                   );

----------------------------------------------------

insert into dept2 (DEPTNO, DNAME, LOC)
values (10, ‘ACCOUNTING‘, ‘NEW YORK‘);

insert into dept2 (DEPTNO, DNAME, LOC)
values (20, ‘RESEARCH‘, ‘DALLAS‘);

insert into dept2 (DEPTNO, DNAME, LOC)
values (30, ‘SALES‘, ‘CHICAGO‘);

insert into dept2 (DEPTNO, DNAME, LOC)
values (40, ‘OPERATIONS‘, ‘BOSTON‘);

============================================================

create table salgrade2 ( grade number,
                         losal number,
                         hisal number
                       );

--------------------------------------------------

insert into salgrade2 (GRADE, LOSAL, HISAL)
values (1, 700, 1200);

insert into salgrade2 (GRADE, LOSAL, HISAL)
values (2, 1201, 1400);

insert into salgrade2 (GRADE, LOSAL, HISAL)
values (3, 1401, 2000);

insert into salgrade2 (GRADE, LOSAL, HISAL)
values (4, 2001, 3000);

insert into salgrade2 (GRADE, LOSAL, HISAL)
values (5, 3001, 9999);


========================================================

select * from emp2;

select * from dept2;

select * from salgrade2;

=====================================
select * from emp2;        --employee2 员工表
select * from dept2;       --department2 部门表
select * from salgrade2;   -- salary grade2 工资等级表

-----------------------------------------------------
emp2

empno    员工编号
ename    员工姓名
job      工作/工种
mgr      manager上级编号
hiredate 入职日期
sal      salary 工资
comm     奖金/津贴
deptno   部门编号
---------------------

dept2

deptno  部门号
dname   部门名称
loc     所在地
---------------------------
salgrade2

grade                   等级
losal lowest salary     最低工资
hisal high salary       最高工资

=============================================



分组查询:

select
from
where       分组前的条件(不允许出现分组函数)
group by    列1,列2,……
having      分组后的条件(关于分组函数的条件)
order by

分组函数 :
   avg() sum() max() min() count() wm_concat()
   平均数 求和 最大  最小   统计    列转行
   
  select avg(sal),sum(sal),max(sal),min(sal),count(sal) from emp2;
  select avg(comm),sum(comm),count(comm) from emp2;
  
分组函数不计算空值




查询每个部门的平均工资:

  
select deptno,avg(sal)
from emp2
group by deptno;





查询每种工作的最高工资:

select job,max(sal)
from emp2
group by job;
  
  






查询每个部门中每种工作的平均工资:

select deptno,job,avg(sal)
from emp2
group by deptno,job
order by deptno;
      
注意：出现在select中的列，必须出现在group by语句里；select集合包含于group by集合
  









查询平均工资大于2000 的部门：

select deptno,avg(sal)
from emp2
group by deptno
having avg(sal) > 2000;
  
  






  
  
查询平均工资大于2000 的部门信息（号、名称、所在地）：

select d.*,avg(sal)
from emp2 e,dept2 d
where e.deptno = d.deptno
group by d.deptno,d.dname,d.loc
having avg(sal) > 2000
order by d.deptno;









查询有有每一个部门有多少人、人名列表、人数：

select deptno,wm_concat(ename),count(ename)
from emp2
group by deptno;
  




----------------------------------------------------------------------------------------------------------
  
子查询：单行子查询 多行子查询 多列子查询
 

查询与SCOTT同部门的员工信息：


(1)select deptno from emp2 where ename = ‘SCOTT‘;
(2)select * from emp2 where deptno = 20;
  
select * from emp2 
where deptno = (select deptno from emp2 where ename = ‘SCOTT‘);
  
  
  
查询与JONES 同上级的员工:


select * from emp2
where mgr = (select mgr from emp2 where ename = ‘JONES‘);


 
查询工资比MILLER 低，奖金比ALLEN 高的员工信息:


select *
from emp2
where sal < (select sal from emp2 where ename = ‘MILLER‘) and comm > (select comm from emp2 where ename = ‘ALLEN‘);
   
   
   
查询与MARTIN 同工作，并且在1981年5 月之前入职的员工:

select *
from emp2
where job = (select job from emp2 where ename = ‘MARTIN‘) and hiredate < to_date(‘1981-5-1‘,‘YYYY-MM-DD‘);
   
   

查询上级是JONES 的员工信息:

select * from emp2
where mgr = (select empno from emp2 where ename = ‘JONES‘);
  


查询工资比平均工资高的员工:

select * from emp2
where sal > (select avg(sal) from emp2);
 


查询工资比10 部门平均工资高的员工:

select * from emp2
where sal > (select avg(sal) from emp2 where deptno = 10);
 


使用子查询，查询SALES 部门的员工信息:

select * 
from emp2 e,dept2 d
where e.deptno = d.deptno and d.dname = ‘SALES‘;

   
select * from emp2 
where deptno = (select deptno from dept2 where dname = ‘SALES‘);













------------------------------------------------------------------------------------------
   








 
多行子查询:   >all    <all    >any     <any     =any     in()
 
查询工资比30 部门所有员工工资都要高的员工信息:

select * from emp2
where sal >all (select sal from emp2 where deptno = 30);


select * from emp2
where sal > (select max(sal) from emp2 where deptno = 30);


  
  >all  大于最大
  
  <all  小于最小
  
  >any  大于最小
  
  <any  小于最大
  
  =any
  
  in()
  
  
  
select * from emp2 where sal in(select sal from emp2 where deptno = 30);

select * from emp2 where sal in(1600,1250,2850,1500,950);



查询哪个部门没有员工:存在于dept表，但是不存在于emp表


select * from dept2
where deptno not in(select distinct deptno from emp2);
  
             select * from emp2;  --deptno 不能有空值
  






多列查询举例：

select * from emp2
where (sal,job) = (select sal,job from emp2 where ename = ‘SCOTT‘ );





  
查询每个部门的部门信息和部门人数(考虑40 部门)：

select d.*,count(e.empno)
from emp2 e right join dept2 d
on e.deptno = d.deptno
group by d.deptno,d.dname,d.loc;
 


select d.deptno,nvl(con,0)
from (select deptno,count(empno) con from emp2 group by deptno) t right join dept2 d
on t.deptno = d.deptno;
 
 
 
 
 
 
P237-238 rownum
 
 
不使用组函数，查询最高工资：


rownum 存在查询结果中，不属于任何一张表
rownum 针对查询结果排序，
        序号从1开始
rownum 比较运算符  = 1   <   <=
 
select * from (select * from emp2 order by sal desc) t
where rownum = 1;











查询工资第二高到第八高的员工信息：
 
(1)
 
 select * from (select * from emp2 order by sal desc) t
 where rownum <= 8
 minus
 select * from (select * from emp2 order by sal desc) t
 where rownum = 1; 
 
 
(2)
-- select t.*,rownum r from ( select * from emp2 order by sal desc  ) t ；

select w.*  from ( select t.*,rownum  r from ( select * from emp2 order by sal desc  ) t ) w where  r between 2 and 8;







   
 3)排序函数
 
 select rank() over(order by sal desc) ran,
        dense_rank() over(order by sal desc) dran,
        row_number() over(order by sal desc) rown,
  emp.*
 from emp;