Oracle row_number分析函数
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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