Oracle 分析函數(shù) row_number(),返回一個整數(shù)值(>=1);
語法格式:
1.row_number() over (order by col_1[,col_2 ...])
作用:按照col_1[,col_2 ...]排序,返回排序后的結(jié)果集,
此用法有點像rownum,為每一行返回一個不相同的值:
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
如果沒有partition by子句, 結(jié)果集將是按照order by 指定的列進行排序;
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;
正如我們所期待的,row_number()返回按照b列排序的結(jié)果,
然后再按照a進行排序,才得到下面的結(jié)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 ...進行分組,
再在每個分組中按照col_1[,col_2 ...]進行排序(升序),
最后返回排好序后的結(jié)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;
這個例子中,我們先按照c列分組,分為2組('*'組,'+'組),
再按照每個小組的b列進行排序(按字符串首字母的ascii碼排),
最后按照a列排序,得到下面的結(jié)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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