oracle累计求和

 

//将当前行某列的值与前面所有行的此列值相加,即累计求和:   

//方法一：   

with t as(    

     select 1 val from dual union all    

     select 3 from dual union all    

     select 5 from dual union all    

     select 7 from dual union all    

     select 9 from dual)    

select val,    

       sum(val)    

       over (order by rownum rows between unbounded preceding and current row)    

       sum_val    

from t    

group by rownum,val    

order by rownum;    

       VAL    SUM_VAL    

---------- ----------    

         1          1    

         3          4    

         5          9    

         7         16    

         9         25    

//解析：    

//sum(val)计算累积和;    

//order by rownum 按照伪列rownum对查询的记录排序;    

//between unbounded preceding and current row:定义了窗口的起点和终点;    

//unbounded preceding:窗口的起点包括读取到的所有行;    

//current row:窗口的终点是当前行，默认值，可以省略;   

//  

//方法二:  

with cte_1 as(    

     select 1 val from dual union all    

     select 3 from dual union all    

     select 5 from dual union all    

     select 7 from dual union all    

     select 9 from dual  

     )   

,cte_2 as(  

    select rownum rn,val from cte_1  

    )  

select a.val , sum(b.val) sum_val  

from cte_2 a , cte_2 b  

where b.rn <= a.rn  

group by a.val  

/  

//方法三：  

//创建一个递归函数，求和  

//f(n) = x + f(n-1)  

create table t  

as  

select 1 id,1 val from dual union all  

select 2,3 from dual union all  

select 3,5 from dual union all  

select 4,7 from dual union all  

select 5,9 from dual  

/  

create or replace function fun_recursion(x in int)   

return integer is  

       n integer :=0;  

begin  

     select val into n   

     from t  

     where id=x;  

     if x=1 then  

        return n;  

     else  

         return n + fun_recursion(x-1);  

     end if;  

     exception  

     when others then  

          dbms_output.put_line(sqlerrm);  

end fun_recursion;  

/  

select val,fun_recursion(id) sum_val from t;  

       VAL    SUM_VAL  

---------- ----------  

         1          1  

         3          4  

         5          9  

         7         16  

         9         25  

//