Question

我有以下递归查询，它将在11g中运行但在Oracle 10g数据库中不受支持：

with st as (
    select
        rownum id,
        case  
            when rownum-1 < 1 then null
            else rownum-1
        end parent_id,
        customer, 
        tickets
    from tickets
),
st2(id, parent_id, customer, tickets, offset) as (
        select
            id, parent_id, shuffler_id, subno, contrno, tickets, 0 offset
        from st
        where id = 1
    union all
        select
            st2.id, st2.parent_id, st2.tickets, (st.tickets + st.offset) offset
        from st, st2
        where st2.parent_id = st.id
)
select * from st2

我要做的是根据以前的行列tickets和offset（同一计算列）计算每行的偏移量，然后我开始第一行偏移量为0;事实上，我需要依赖于我在查询执行期间计算的列需要递归。

问题是Oracle 10g不支持上述查询，所以我尝试使用connect by而不是有效，但丑陋的部分是效率极低：

with st as (
    select
        rownum id,
        case  
            when rownum-1 < 1 then null
            else rownum-1
        end parent_id, 
        customer,
        tickets
    from tickets
)
select
    id, 
    parent_id,
    customer,
    tickets,
    (
        select nvl(sum(tickets), 0) from st
        where level <  x.id
        start with id = 1
        connect by prior id = parent_id
    ) offset
from st x

使用第二个查询我总结了所有以前的行，这些行完成了工作但也是多余的，我不能依赖于这个表何时增长到数百万条记录。

如何实现类似于Oracle 10g数据库中第一个查询的东西？

Answer 1

  select 
    id, 
    customer,
    tickets,
    nvl(
      sum(tickets) over (
        order by id rows between unbounded preceding and 1 preceding
      ), 0) as offset
  from (    
    select
        rownum id,
        customer, 
        tickets
    from tickets
  )

或更短（不引入id s）

  select 
    customer,
    tickets,
    nvl(
      sum(tickets) over (
        order by rownum rows between unbounded preceding and 1 preceding
      ), 0) as offset
  from tickets

Oracle 10g递归查询

1 个答案: