Oracle row_number分析函数

Oracle 分析函数 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

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