Question

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。

以下是数据：

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);

戴夫引用了海伦领导的树。

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;

但是结果不包括海伦的树：

爱丽丝，辛迪，盖尔，弗雷德，鲍勃，戴夫，埃玛

我如何获得海伦树的完整结果：

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：

爱丽丝，鲍勃，戴夫，海伦，伊戈尔，杰罗姆，埃玛，辛迪，弗雷德，盖尔

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

因为您要针对同一案例要求不同的结果：

Cindy和Bob：同一个父母。辛迪（Cindy）和鲍勃（Bob）同胞，而鲍勃（Bob）之前你要辛迪（Cindy）。
Jerome和Igor：同一个父母。杰罗姆（Jerome）与伊戈尔（Igor）同胞，而你想让伊戈尔（Igor）在杰罗姆（Jerome）前。

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

我认为预期输出应为：

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

OR

爱丽丝，鲍勃，戴夫，海伦，伊戈尔，杰罗姆，埃玛，辛迪，弗雷德，盖尔

第一种方法

我尝试了以下查询（使用refs.id代替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;

爱丽丝，鲍勃，戴夫，海伦，伊戈尔，杰罗姆，埃玛，辛迪，弗雷德，盖尔

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

爱丽丝，鲍勃，戴夫，海伦，伊戈尔，杰罗姆，埃玛，辛迪，弗雷德，盖尔

Answer 2

这将起作用：

const searchArea = `s-kreuzberg`
const searchParam = `bike`
const url = `https://www.ebay-kleinanzeigen.de/${searchArea}/seite:1/${searchParam}/k0l3375r5`

async function go() {
  await getHTML(url)
}

go()

必需的输出：

if not 'f' in df.columns:
    df = df.withColumn('f', <>)

With子句：嵌套树

2 个答案:

第一种方法

第二种方法