Parttion by 关键字是Oracle中分析性函数的一部分，它和聚合函数不同的地方在于它能够返回一个分组中的多条记录，儿聚合函数一般只有一条反映统计值的结果。
场景：查询出每个部门工资最低的员工编号【每个部门可能有两个最低的工资员工】
create table TSALER
(
userid NUMBER(10),
salary NUMBER(10),
deptid NUMBER(10)
)
-- Add comments to the columns
comment on column TSALER.userid
is ‘员工ID‘;
comment on column TSALER.salary
is ‘工资‘;
comment on column TSALER.deptid
is ‘部门ID‘;
insert into TSALER (工号, 工资, 部门编号)
values (1, 200, 1);
insert into TSALER (工号, 工资, 部门编号)
values (2, 2000, 1);
insert into TSALER (工号, 工资, 部门编号)
values (3, 200, 1);
insert into TSALER (工号, 工资, 部门编号)
values (4, 1000, 2);
insert into TSALER (工号, 工资, 部门编号)
values (5, 1000, 2);
insert into TSALER (工号, 工资, 部门编号)
values (6, 3000, 2);
查询结果：
select tsaler.* from tsaler
inner join(select min(salary) as salary,deptid from tsaler group by deptid) c
on tsaler.salary=c.salary and tsaler.deptid=c.deptid
select * from tsaler
inner join(select min(salary) as salary,deptid from tsaler group by deptid) c
using(salary,deptid)
--row_number() 顺序排序
select row_number() over(partition by deptid order by salary) my_rank ,deptid,USERID,salary from tsaler;
--rank() （跳跃排序，如果有两个第一级别时，接下来是第三级别）
select rank() over(partition by deptid order by salary) my_rank,deptid,USERID,salary from tsaler;
--dense_rank（）（连续排序，如果有两个第一级别时，接下来是第二级）
select dense_rank() over(partition by deptid order by salary) my_rank,deptid,USERID,salary from tsaler;
-------方案3解决方案
select * from (select rank() over(partition by deptid order by salary) my_rank,deptid,USERID,salary from tsaler) where my_rank=1;
select * from (select dense_rank() over(partition by deptid order by salary) my_rank,deptid,USERID,salary from tsaler) where my_rank=1;
原文：http://www.cnblogs.com/lcword/p/5719303.html