oracle row_number分析函数

oracle 分许函数 row_number(),返回一个整数值(>=1);

语法格式:

1.row_number() over (order by col_1[,col_2 ...])

作用:按照col_1[,col_2 ...]排序,返回排序后的结果集，

此用法有点像rownum,为每一行返回一个不相同的值：

select rownum,ename,job, row_number() over (order by rownum) row_number from emp; ROWNUM ENAME JOB ROW_NUMBER ---------- ---------- --------- ---------- 1 SMITH CLERK 1 2 ALLEN SALESMAN 2 3 WARD SALESMAN 3 4 JONES MANAGER 4 5 MARTIN SALESMAN 5 6 BLAKE MANAGER 6 7 CLARK MANAGER 7 8 SCOTT ANALYST 8 9 KING PRESIDENT 9 10 TURNER SALESMAN 10 11 ADAMS CLERK 11 12 JAMES CLERK 12 13 FORD ANALYST 13 14 MILLER CLERK 14

如果没有partition by子句, 结果集将是按照order by 指定的列进行排序；

with row_number_test as( select 22 a,'twenty two' b from dual union all select 1,'one' from dual union all select 13,'thirteen' from dual union all select 5,'five' from dual union all select 4,'four' from dual) select a,b, row_number() over (order by b) from row_number_test order by a;

正如我们所期待的,row_number()返回按照b列排序的结果,

然后再按照a进行排序,才得到下面的结果:

A B ROW_NUMBER()OVER(ORDERBYB) ---------- ---------- -------------------------- 1 one 3 4 four 2 5 five 1 13 thirteen 4 22 twenty two 5

2.row_number() over (partition by col_n[,col_m ...] order by col_1[,col_2 ...])

作用:先按照col_n[,col_m ...进行分组,

再在每个分组中按照col_1[,col_2 ...]进行排序(升序),

最后返回排好序后的结果集:

with row_number_test as( select 22 a,'twenty two' b,'*' c from dual union all select 1,'one','+' from dual union all select 13,'thirteen','*' from dual union all select 5,'five','+' from dual union all select 4,'four','+' from dual) select a,b, row_number() over (partition by c order by b) row_number from row_number_test order by a;

这个例子中，我们先按照c列分组，分为2组('*'组,'+'组)，

再按照每个小组的b列进行排序(按字符串首字母的ascii码排),

最后按照a列排序,得到下面的结果集:

A B ROW_NUMBER ---------- ---------- ---------- 1 one 3 4 four 2 5 five 1 13 thirteen 1 22 twenty two 2

原文：http://www.adp-gmbh.ch/ora/sql/analytical/row_number.html