oracle row_number()

要求查询每个用户对应的最大样品信息，忽然想到ms sql提供过 row_number() over(partition by 列 order by 列 desc），那么oracle可能也存在，

我的表结构如下：

create table NEOGOODSRULE
(
  ID          NUMBER(22) not null,
  PERSONALID  NVARCHAR2(50),
  CT_SMP_TYPE NVARCHAR2(100)
)
tablespace VGSM
  pctfree 10
  initrans 1
  maxtrans 255
  storage
  (
    initial 64K
    minextents 1
    maxextents unlimited
  );

数据如下：

insert into neogoodsrule (ID, PERSONALID, CT_SMP_TYPE)
values (2270, ‘JYZ‘, ‘原料‘);

insert into neogoodsrule (ID, PERSONALID, CT_SMP_TYPE)
values (2271, ‘JYZ‘, ‘辅料‘);

insert into neogoodsrule (ID, PERSONALID, CT_SMP_TYPE)
values (2359, ‘SYSTEM‘, ‘包材(内)‘);

insert into neogoodsrule (ID, PERSONALID, CT_SMP_TYPE)
values (2360, ‘SYSTEM‘, ‘包材(外)‘);

insert into neogoodsrule (ID, PERSONALID, CT_SMP_TYPE)
values (2361, ‘SYSTEM‘, ‘原料‘);

insert into neogoodsrule (ID, PERSONALID, CT_SMP_TYPE)
values (2362, ‘SYSTEM‘, ‘成品‘);

insert into neogoodsrule (ID, PERSONALID, CT_SMP_TYPE)
values (2363, ‘SYSTEM‘, ‘稳定性(加速)‘);

insert into neogoodsrule (ID, PERSONALID, CT_SMP_TYPE)
values (2364, ‘SYSTEM‘, ‘稳定性(长期)‘);

insert into neogoodsrule (ID, PERSONALID, CT_SMP_TYPE)
values (2365, ‘SYSTEM‘, ‘辅料‘);

insert into neogoodsrule (ID, PERSONALID, CT_SMP_TYPE)
values (2354, ‘LY‘, ‘成品‘);

insert into neogoodsrule (ID, PERSONALID, CT_SMP_TYPE)
values (2355, ‘LY‘, ‘原料‘);

ROW_NUMBER（）语法如下：

1、row_number() over(order by column asc)先对列column按照升序，再为每条记录返回一个序列号：

select personalid,row_number() over(order by personalid asc) rn from  neogoodsrule

2、row_number() over(partition by column1 order by column2 asc) 先按照column1分组，再对分组后的数据进行以column2升序排列

select personalid,ct_smp_type,row_number() over(partition by personalid order by ct_smp_type asc) rn from neogoodsrule

由此，开始所提的需求sql代码如下

select * from (select personalid,ct_smp_type,row_number() over(partition by personalid order by ct_smp_type asc) rn from neogoodsrule )
where rn=1

使用row_number()分页函数取代group by。例如：语句1:
select col1,col2 from t where col1 in (select col1 from t group by col1 having count(*) <=2 ) order

by order col2;
语句1可以用下面的语句取代：
语句2:
select col1,col2 from (select *,row_number() over(partition by col1 order by col2) as p_group from

t) where p_group<=2 ;

ROW_NUMBER() OVER (PARTITION BY COL1 ORDER BY COL2)
表示根据COL1分组，在分组内部根据 COL2排序。而这个值就表示每组内部排序后的顺序编号（组内连续的唯一的

）。

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

select a,b,c,sum(c)over(partition by a) from t2                
   得到结果：
   A   B   C        SUM(C)OVER(PARTITIONBYA)      
   -- -- ------- ------------------------ 
   h   b   3        3                        
   m   a   2        4                        
   m   a   2        4                        
   n   a   3        6                        
   n   b   2        6                        
   n   b   1        6                        
   x   b   3        9                        
   x   b   2        9                        
   x   b   4        9                        
  
   如果用sum，group by 则只能得到
   A   SUM(C)                            
   -- ---------------------- 
   h   3                      
   m   4                      
   n   6                      
   x   9                      
   无法得到B列值

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

RANK() OVER (PARTITION BY COL1 ORDER BY COL2)
类似，不过RANK 排序的时候跟派名次一样，可以并列2个第一名之后 是第3名。

dense_rank() OVER (PARTITION BY COL1 ORDER BY COL2)
类似。是连续排序，有两个第二名时仍然跟着第三名。

dept sal emp row_number rank dense_rank
---- ------- ----- ---------- ----- ----------
10 5000.00 7839 1 1 1
10 2450.00 7782 2 2 2
10 1300.00 7934 3 3 3
20 3000.00 7788 1 1 1
20 3000.00 7902 2 1 1
20 2975.00 7566 3 3 2
20 1100.00 7876 4 4 3
20 800.00 7369 5 5 4
30 2850.00 7698 1 1 1
30 1600.00 7499 2 2 2

在求第一名成绩的时候，不能用row_number()，因为如果同班有两个并列第一，row_number()只返回一个结果