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oracle row_number()

2024-08-15 06:36:21   211人阅读

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

oracle row_number()


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