一次查询 SQLite 选择所有递归值答案

【问题标题】：Select all recursive value with one query SQLite一次查询 SQLite 选择所有递归值
【发布时间】：2014-07-21 15:12:54
【问题描述】：

我还有一个案例：

首先我创建了一个表：

 CREATE TABLE tree(
     id_tree integer PRIMARY KEY AUTOINCREMENT,
     id_boss TEXT,
     id_child TEXT,
     answ TEXT);

插入一些值：

 INSERT INTO tree(id_boss,id_child,answ) VALUES('1','8','T');
 INSERT INTO tree(id_boss,id_child,answ) VALUES('1',null,'F');
 INSERT INTO tree(id_boss,id_child,answ) VALUES('8','P1','T');
 INSERT INTO tree(id_boss,id_child,answ) VALUES('8','2','F');
 INSERT INTO tree(id_boss,id_child,answ) VALUES('2','P2','T');
 INSERT INTO tree(id_boss,id_child,answ) VALUES('2','P3','F');

并执行查询：

  WITH RECURSIVE
  ancestors(id, answ) AS (
  VALUES('P3', 'T')
  UNION ALL
  SELECT tree.id_boss, tree.answ
  FROM tree JOIN ancestors ON tree.id_child = ancestors.id
  )
  SELECT id FROM ancestors WHERE answ = 'T';

结果是：

  P3
  1

对于P3，我想列出所有递归值，所以它会是这样的：

   1 --- // P3
   1 --- // P1
   8 --- // P1
   2 --- // P2
   1 --- // P2

【问题讨论】：

标签： sql recursion

【解决方案1】：

这是您要找的附近的某个地方吗？

WITH seed (id, answ) as ( VALUES('P3', 'T') )
   , ancestors1(id, answ) AS (
        select * from seed
        UNION ALL
        SELECT tree.id_boss, tree.answ
        FROM tree, ancestors1 where tree.id_child = ancestors1.id
     )
   , ancestors2(id, answ) AS (
        select * from seed
        UNION ALL
        SELECT tree.id_boss, tree.answ
        FROM tree, ancestors2 where tree.id_child = ancestors2.id
    )
select id from (
    select * from ancestors1
    union all
    select * from ancestors2
);

我不得不改写某些东西，以便让他们用 db2 编译（删除 RECURSIVE，并使用隐式连接）

【讨论】：

对不起，但不起作用，我想选择所有递归 = P1，P2，P3，而不仅仅是 P3 值。