oracle累計求和
//將當前行某列的值與前面所有行的此列值相加,即累計求和:
//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 13 4
5 97 16
9 25
//解析：
//sum(val)計算累積和;
//order by rownum 按照偽列rownum對查詢的記錄排序;
//between unbounded preceding and current row:定義了視窗的起點和終點;
//unbounded preceding:視窗的起點包括讀取到的所有行;
//current row:視窗的終點是當前行，預設值，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：
//建立乙個遞迴函式，求和
//f(n) = x + f(n-1)
create table t
asselect 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 13 4
5 97 16
9 25
參考文件
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