我有一个具有同级排序的树层次结构。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);
戴夫(Dave)引用了海伦(Helen)领导的那棵树。
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;
但是结果不包括海伦的树:
我怎样才能得到海伦的树的完整结果:
编辑:
鲍勃(Bob)和辛迪(Cindy)以及弗雷德＆盖尔(Fred＆Gail)被颠倒了……实际预期结果是:
最佳答案
我认为您无法获得以下预期输出:
因为您要针对同一案例寻求不同的结果:
(如果必须在Bob之前打印Cindy，那么必须在Igor之前打印Jerome)
我认为预期的输出应为:
或者
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;
第二种方法
我使用了另一种方法:
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
关于sql - 有子句: nesting trees，我们在Stack Overflow上找到一个类似的问题：https://stackoverflow.com/questions/55549902/