Question

我有一个递归查询，它确实扩展了这个Java猴子SQL知识的限制。现在它终于凌晨1点半了，现在可能是时候开始寻求帮助了。这是谷歌失败的少数几次之一。

表格如下：

Parent_ID CHILD_ID QTY
25        26        1
25        27        2
26        28        1
26        29        1
26        30        2
27        31        1
27        32        1
27        33        2

我正在尝试获得以下结果，其中父项列出了下面的每个子项。请注意qty的级联。

BASE    PARENT_ID  CHILD_ID   QTY
25         25        26        1
25         25        27        2
25         26        28        1
25         26        29        1
25         26        30        1
25         27        31        2
25         27        32        2
25         27        33        4
26         26        28        1
26         26        29        1
26         26        30        2
27         27        31        1
27         27        32        1
27         27        33        2

我尝试了以下几种偏差但无济于事。

SELECT *
FROM MD_BOMS
START WITH PARENT_ID is not null
CONNECT BY PRIOR CHILD_ID = PARENT_ID
ORDER BY PARENT_ID, CHILD_ID

我正在使用Oracle数据库。任何建议，想法等将不胜感激。这似乎很接近，但我不确定这是否是我要找的：Retrieve all Children and their Children, recursive SQL

基于（Retrieve all Children and their Children, recursive SQL）我也尝试过以下操作，但收到“非法引用WITH子句中的查询名称”错误：

with cte as (
    select  CHILD_ID, PARENT_ID, CHILD_ID as head
    from    MD_BOMS
    where   PARENT_ID is not null
    union all
    select  ch.CHILD_ID, ch.PARENT_ID, p.head
    from    MD_BOMS ch
    join    cte pa
    on      pa.CHILD_ID = ch.PARENT_ID
)
select  *
from    cte

Answer 1

你很近：

select connect_by_root parent_id base, parent_id, child_id, qty
from md_boms
connect by prior child_id = parent_id
order by base, parent_id, child_id;

          BASE  PARENT_ID   CHILD_ID        QTY
    ---------- ---------- ---------- ----------
            25         25         26          1 
            25         25         27          2 
            25         26         28          1 
            25         26         29          1 
            25         26         30          2 
            25         27         31          1 
            25         27         32          1 
            25         27         33          2 
            26         26         28          1 
            26         26         29          1 
            26         26         30          2 
            27         27         31          1 
            27         27         32          1 
            27         27         33          2 

     14 rows selected

connect_by_root operator为您提供基础parent_id。

SQL Fiddle

我不确定你是如何计算qty的。我猜你想要孩子的路径总数，但这与你所展示的不符。作为一个起点，然后，从this answer大量借用非常，你可以尝试类似的东西：

with hierarchy as (
  select connect_by_root parent_id base, parent_id, child_id, qty,
    sys_connect_by_path(child_id, '/') as path
  from md_boms
  connect by prior child_id = parent_id
)
select h.base, h.parent_id, h.child_id, sum(e.qty)
from hierarchy h
join hierarchy e on h.path like e.path ||'%'
group by h.base, h.parent_id, h.child_id
order by h.base, h.parent_id, h.child_id;

     BASE  PARENT_ID   CHILD_ID SUM(E.QTY)
---------- ---------- ---------- ----------
        25         25         26          1 
        25         25         27          2 
        25         26         28          2 
        25         26         29          2 
        25         26         30          3 
        25         27         31          3 
        25         27         32          3 
        25         27         33          4 
        26         26         28          1 
        26         26         29          1 
        26         26         30          2 
        27         27         31          1 
        27         27         32          1 
        27         27         33          2 

 14 rows selected

Answer 2

@AlexPoole答案很棒，我只想用更直观的查询方式扩展他的答案，以便沿路径求和值。
此变体基于Oracle 11g R2中引入的recursive subquery factoring功能。

with recursion_view(base, parent_id, child_id, qty) as (
   -- first step, get rows to start with
   select 
     parent_id base, 
     parent_id, 
     child_id, 
     qty
  from 
    md_boms

  union all

  -- subsequent steps
  select
    -- retain base value from previous level
    previous_level.base,
    -- get information from current level
    current_level.parent_id,
    current_level.child_id,
    -- accumulate sum 
    (previous_level.qty + current_level.qty) as qty 
  from
    recursion_view previous_level,
    md_boms        current_level
  where
    current_level.parent_id = previous_level.child_id

)
select 
  base, parent_id, child_id, qty
from 
  recursion_view
order by 
  base, parent_id, child_id

SQLFiddle example（扩展了一个数据行以演示超过2个级别的工作）

如何在Oracle SQL中检索父行的所有递归子级？

2 个答案: