With 子句：嵌套树

Question

我有一个带有兄弟顺序的树状层次结构。我需要添加对其他树的引用。

这是数据：

drop table if exists org; CREATE TABLE org(id int primary key, name text, boss int, sibling int, ref int) without rowid;
INSERT INTO org VALUES(0, 'Alice', NULL, null, null);
INSERT INTO org VALUES(1, 'Bob', 0, null, null);
INSERT INTO org VALUES(2, 'Cindy', 0, 1, null);
INSERT INTO org VALUES(3, 'Dave', 1, 4, 7);
INSERT INTO org VALUES(4, 'Emma', 1, null, null);
INSERT INTO org VALUES(5, 'Fred', 2, null, null);
INSERT INTO org VALUES(6, 'Gail', 2, 5, null);
INSERT INTO org VALUES(7, 'Helen', NULL, null, null);
INSERT INTO org VALUES(8, 'Igor', 7, null, null);
INSERT INTO org VALUES(9, 'Jerome', 7, 8, null);

Dave 引用了 Helen 领导的树。

我添加了 refs 子句：

WITH RECURSIVE
refs(id, name, boss, sibling, ref, lref) AS (
 SELECT id, name, boss, sibling, ref, 0 FROM org
UNION ALL
 SELECT org.id, org.name, org.boss, org.sibling, org.ref, refs.lref+1
 FROM org JOIN refs ON org.id=refs.ref
),
sibs(id, name, boss, lref, lsib) AS (
 SELECT id, name, boss, lref, 0 FROM refs
 WHERE sibling IS NULL
UNION ALL
 SELECT refs.id, refs.name, refs.boss, refs.lref, sibs.lsib + 1
 FROM refs
 JOIN sibs ON refs.boss = sibs.boss
 AND refs.sibling = sibs.id
),
tree(id, name, lsib, lref, level) AS (
 select id, name, 0, 0, 0 from org where id = 0
UNION ALL
 SELECT sibs.id, sibs.name, sibs.lsib, sibs.lref, tree.level+1
 FROM sibs JOIN tree ON sibs.boss=tree.id
ORDER BY 4 DESC, 5 DESC, 3 DESC
)
SELECT group_concat(name) FROM tree;

但结果不包括海伦的树：

Alice,Cindy,Gail,Fred,Bob,Dave,Emma

我怎样才能得到海伦树的完整结果:

Alice,Cindy,Gail,Fred,Bob,Dave,Helen,Igor,Jerome,Emma

编辑：

Bob 和 Cindy - 以及 Fred&Gail- 被颠倒了......实际预期结果是：

Alice,Bob,Dave,Helen,Igor,Jerome,Emma,Cindy,Fred,Gail

Answer 1

我认为您无法获得以下预期输出：

Alice,Cindy,Gail,Fred,Bob,Dave,Helen,Igor,Jerome,Emma

因为你要求对同一个案例有不同的结果：

Cindy 和 Bob：同父异母。 Cindy 是 Bob 的兄弟姐妹，你想要 Cindy 而不是 Bob。
杰罗姆和伊戈尔：同父异母。杰罗姆是伊戈尔的兄弟姐妹，你想要伊戈尔而不是杰罗姆。

（如果 Cindy 必须在 Bob 之前打印，那么 Jerome 必须在 Igor 之前打印）

我认为预期的输出应该是：

Alice,Cindy,Gail,Fred,Bob,Dave,Helen,Jerome,Igor,Emma

或

Alice,Bob,Dave,Helen,Igor,Jerome,Emma,Cindy,Fred,Gail

第一种方法

我尝试了以下查询 （使用 refs.id 而不是 org.boss）:

WITH RECURSIVE
refs(id, name, boss, sibling, ref, lref) AS (
 SELECT id, name, boss, sibling, ref, 0 FROM org

UNION ALL
 SELECT org.id, org.name, refs.id, org.sibling, org.ref, refs.lref+1
 FROM org JOIN refs ON org.id=refs.ref
),
sibs(id, name, boss, lref, lsib,ref) AS (
 SELECT id, name, boss, refs.lref, 0,ref FROM refs
  WHERE sibling IS NULL
UNION ALL
 SELECT refs.id, refs.name, refs.boss, refs.lref, sibs.lsib + 1,refs.ref
 FROM refs
 JOIN sibs ON refs.boss = sibs.boss
 AND refs.sibling = sibs.id
),
tree(id, name, lsib, lref, level) AS (
 select org.id, org.name, 0, 0, 0 from org 
  where id = 0

UNION ALL
 SELECT sibs.id, sibs.name, sibs.lsib, sibs.lref, tree.level+1
 FROM sibs JOIN tree ON sibs.boss = tree.id 
ORDER BY  4 DESC, 5 DESC
)
select group_concat(name) from tree;

Alice,Bob,Dave,Helen,Igor,Jerome,Emma,Cindy,Fred,Gail

第二种方法

我使用了不同的方法：

WITH RECURSIVE
pc(id,name,parent,priority) AS(
  select org.id,org.name,coalesce(refs.id,org.boss,-1) as "parent",
  case
  when refs.id is not null then 3
  when sibls.id is not null then 2
  when org.boss is not null then 1
  else 0 end as  "priority"
  from org left join org as refs on org.id = refs.ref
  left join org as sibls on org.id = sibls.sibling
where org.id > 0),
  tree(id,name,parent,priority,level) AS(
    select id,name,0,0,0 from org where id = 0
    UNION ALL
    select pc.id,pc.name,pc.parent,pc.priority,tree.level + 1 from pc 
    join tree on tree.id = pc.parent
    order by 3 desc )
    select group_concat(name) from tree

Alice,Bob,Dave,Helen,Igor,Jerome,Emma,Cindy,Fred,Gail

Answer 2

这会起作用：

SELECT LISTAGG(name, ', ') WITHIN GROUP (ORDER BY name)
      FROM org
      START WITH boss is null 
      CONNECT BY PRIOR id= boss
      ORDER SIBLINGS BY sibling;

要求输出：

Alice, Bob, Cindy, Dave, Emma, Fred, Gail, Helen, Igor, Jerome

With 子句：嵌套树

With clause : nesting trees

sql

sqlite

common-table-expression

第一种方法

第二种方法