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sql - 带有子句 : nesting trees

转载 作者:IT王子 更新时间:2023-10-29 06:26:43 26 4
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我有一个具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

最佳答案

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 sibling ,您希望 Cindy 优先于 Bob。
  • 杰罗姆和伊戈尔:同父异母。杰罗姆是伊戈尔的 sibling ,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

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

关于sql - 带有子句 : nesting trees,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/55549902/

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