使用 MySQL 8.0 递归 CTE 查找直系后代的数量并传播到分层 table 中的父代

Question

我试图在层次结构中的每个节点上累积直系后代的数量。对于没有后代的节点，计数应为 0。

总的来说，我想在多个上下文中应用不同类型的 counting/aggregation，在这些上下文中，层次结构没有像这样准确定义。因此我对递归解决方案很感兴趣。

考虑下面的代码。我如何 "invert" 这个查询，而不是计算深度，我计算后代并向上传播数字？

create table hierarchy (
    name     varchar(100),
    location varchar(100),
    parent_name varchar(100),
    parent_location varchar(100)
) engine=InnoDB default charset=UTF8MB4;

truncate hierarchy;
insert into hierarchy values
    ('music', '/', NULL, NULL),
    ('classical', '/music', 'music', '/'),
    ('pop', '/music', 'music', '/'),
    ('rock', '/music', 'music', '/'),
    ('bach', '/music/classical', 'classical', '/music');
select * from hierarchy;

with recursive cte as
(
    select name, location, parent_name, parent_location, 1 as depth
    from hierarchy where parent_name is NULL and parent_location is NULL
    union all
    select a.name, a.location, a.parent_name, a.parent_location, depth + 1
    from hierarchy as a inner join cte on a.parent_name = cte.name and a.parent_location = cte.location
)
select *
from cte;

输出为

name         location           parent_name   parent_location   depth
'music'      '/'                NULL          NULL              1
'classical'  '/music'           'music'       '/'               2
'pop'        '/music'           'music'       '/'               2
'rock'       '/music'           'music'       '/'               2
'bach'       '/music/classical' 'classical'   '/music'          3

我最终感兴趣的是这个输出：

name         location           parent_name   parent_location   descendents
'music'      '/'                NULL          NULL              3
'classical'  '/music'           'music'       '/'               1
'pop'        '/music'           'music'       '/'               0
'rock'       '/music'           'music'       '/'               0
'bach'       '/music/classical' 'classical'   '/music'          0

Answer 1

您似乎想要计算每个节点的直系后代的数量。如果是这样，我认为您不需要递归查询：一个简单的子查询就可以做到：

select 
    h.*,
    (select count(*) from hierarchy h1 where h1.parent_name = h.name) descendants
from hierarchy h

Demo on DB Fiddle:

| name      | location         | parent_name | parent_location | descendants |
| --------- | ---------------- | ----------- | --------------- | ----------- |
| music     | /                |             |                 | 3           |
| classical | /music           | music       | /               | 1           |
| pop       | /music           | music       | /               | 0           |
| rock      | /music           | music       | /               | 0           |
| bach      | /music/classical | classical   | /music          | 0           |

使用 MySQL 8.0 递归 CTE 查找直系后代的数量并传播到分层 table 中的父代

Find number of direct descendents and propagate to parents in hierarchical table with MySQL 8.0 recursive CTE

mysql

sql

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recursive-query

hierarchical-data