Postgres中的递归SQL查询

Scema：

示例表：

Person(pid:int,
       name:char(20),
       predecessor:int -> Person.pid
      )

如何找到一个人“abc”的所有继任者

所需输出：

pid, name, predecessor
0,   'abc', NULL
1,   'bcd', 0
2,   'cde', 1
3,   'efg', NULL
4,   'fgh', 3

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非常感谢

您可以通过生成所有祖先，然后过滤掉它们来实现这一点。以下是您的数据示例：

name 
'bcd'
'cde' 

---

是一个SQL FIDLE。

只需在一个：

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旁白：

请编辑问题，提供样本数据和所需结果。检查手册。它有这样的例子：

with recursive cte(pid, lev, ancestor) as (
      select pid, 0, predecessor
      from person p 
      union all
      select cte.pid, lev + 1, p.predecessor
      from person p join
           cte
           on p.pid = cte.ancestor
     )
select p2.name
from cte join
     person p1
     on cte.ancestor = p1.pid join
     person p2
     on cte.pid = p2.pid
where p1.name = 'abc';

with recursive cte as (
   select p.pid, p.name, 1 AS lvl
   from   person a
   join   person p ON p.predecessor = a.pid
   where  a.name = 'abc'

   union all
   select p.pid, p.name, c.lvl + 1
   from   cte    c
   join   person p ON  p.predecessor = c.pid
   )
select name
from   cte
order  by lvl;