获取父母的所有后代

Question

我有一个包含两列的表，Parent并且Child.

我需要获取与父记录关联的所有后代的列表。

Source Table
+----+-----------+
| Parent | Child |
+----+-----------+
|  a     |     b |
|  b     |     c |
|  c     |     d |
|  d     |     e |
|  e     |     f |
|  f     |     x |
+----+-----------+

Expected Result:
+----+-----------+
| Parent | Child |
+----+-----------+
|  a     |     b |  // As b is the child of a, all the descendants of b 
|  a     |     c |  // are also descendants of a. 
|  a     |     d |
|  a     |     e |
|  a     |     f |
|  a     |     x |
|  b     |     c |  // As c is the child of b, all the descendants of c 
|  b     |     d |  // are also descendants of b.
|  b     |     e |
|  b     |     f |
|  b     |     x |
|  c     |     d |
|  c     |     e |
|  c     |     f |
|  c     |     x |
|  d     |     e |
|  d     |     f |
|  d     |     x |
|  e     |     f |
|  e     |     x |
|  f     |     x |
+----+-----------+

任何想法如何获取父母的所有后代记录？我尝试过使用 self join ，T1.Child = T2.Parent但逻辑不起作用。

我正在使用不支持子句的Teiid VDBRecursive WITH。

Answer 1

您需要一个递归 CTE（公用表表达式）：

with    -- recursive
        -- some DBMS (e.g. Postgres) require the word "recursive"
        -- some others (Oracle, SQL-Server) require omitting the "recursive"
        -- and some (e.g. SQLite) don't bother, i.e. they accept both
descendants
  (parent, descendant, lvl) 
as
  ( select parent, child, 1
    from source
  union all
    select d.parent, s.child, d.lvl + 1
    from descendants  d
      join source  s
        on d.descendant = s.parent
  ) 
select *
from descendants 
order by parent, lvl, descendant ;

请参阅SQLfiddle。在level我加入没有必要的。它为您提供父母和后代之间的“距离”（1 是孩子，2 是孙子等）。

相关 Teiid 文档的链接：(recursive) CTEs。

PS：对于甲骨文：

• LEVEL 是受限关键字