带递归的 MySQL

Question

我需要返回我的类别以及每个部门的商品数量，父类别必须从子类别继承每个类别的产品数量。

类别

+-------------+----------+--------+ |编号 |姓名 | parent_id | +-------------+----------+--------+ | 1 |电子产品 |空 | | 2 |电视 | 1 | | 3 |管 | 2 | | 4 |液晶显示器 | 2 | | 5 |等离子 | 2 | | 6 |便携式电子产品 | 1 | | 7 | MP3 播放器 | 6 | | 8 |闪存 | 7 | | 9 | CD 播放器 | 6 | | 10 | 2 路收音机 | 6 | +-------------+----------+--------+

产品

+-------------+----------+--------+ |编号 |产品 |类别 ID | +-------------+----------+--------+ | 1 |液晶电视 32" | 4 | | 2 |液晶电视 45" | 4 | | 3 |电视管 29" | 3 | | 3 | IPOD | 7 | +-------------+----------+--------+

预期结果

+-------------+----------+------------ + |编号 |姓名 |等级|数量 | +-------------+----------+------------ + | 1 |电子产品 | 1 | 4 | | 2 |电视 | 2 | 3 | | 3 |管 | 3 | 1 | | 4 |液晶显示器 | 3 | 2 | | 5 |便携式电子产品 | 2 | 1 | | 6 | MP3 播放器 | 3 | 1 | +-------------+----------+------------ +

我需要使用with递归，因为速度比使用嵌套要快很多

WITH RECURSIVE category_path (id, name, level, parent_id) AS
(
SELECT id, name, 1 level, parent_id
 FROM categories
  WHERE parent_id IS NULL
 UNION ALL
 SELECT c.id, c.name, level + 1, c.parent_id
  FROM category_path AS cp 
   JOIN categories AS c
    ON cp.id = c.parent_id
)
SELECT * FROM category_path

Time: 0.020s

使用嵌套

SELECT
     parent.id,     
   parent.name,
     parent.parent_id,
   COUNT(departaments.product_id) 
FROM
   categories AS node 
   INNER JOIN
      categories AS parent 
      ON node.lft BETWEEN parent.lft AND parent.rgt 
   INNER JOIN
      departaments 
      ON node.id = departaments.categorie_id 
GROUP BY
   parent.id 
ORDER BY
   node.lft;

Time: 1.510s

Answer 1

首先编写一个查询来获取每个类别的产品数量。这很简单：

with products_per_category as (
  select c.id, count(p.id) as pcount
  from categories c
  left join products p on p.category_id = c.id
  group by c.id
)
  select *
  from products_per_category
  order by id

db-fiddle

然后编写递归CTE，生成transitive closure：

with recursive rcte as (
  select c.id, c.id as ancestor_id
  from categories c
  union all
  select r.id, c.parent_id
  from rcte r
  join categories c on c.id = r.ancestor_id
)
  select *
  from rcte
  order by id, ancestor_id

结果会是这样的：

| id  | ancestor_id |
| --- | ----------- |
| 1   | 1           |
| 2   | 1           |
| 2   | 2           |
...
| 9   | 1           |
| 9   | 6           |
| 9   | 9           |
| 10  | 1           |
| 10  | 6           |
| 10  | 10          |

db-fiddle

这就像你得到从根节点到每个类别的路径。例如。对于9，路径是1->6->9

如果您先通过ancestor_id订购，您将获得：

| id  | ancestor_id |
| --- | ----------- |
| 1   | 1           |
...
| 10  | 1           |
| 2   | 2           |
| 3   | 2           |
| 4   | 2           |
| 5   | 2           |
| 3   | 3           |
...

db-fiddle

在这里您可以看到，类别 2 (ancestor_id=2) 具有子类别 (id) 2,3,4,5。请注意，每个类别都有自己的子类别。这将使下一步更简单。

现在我们只需要加入两个 CTE 并总结产品数量：

with recursive products_per_category as (
  select c.id, count(p.id) as pcount
  from categories c
  left join products p on p.category_id = c.id
  group by c.id
), rcte as (
  select c.id, c.id as ancestor_id
  from categories c
  union all
  select r.id, c.parent_id
  from rcte r
  join categories c on c.id = r.ancestor_id
  where c.parent_id is not null
)
  select
    c.id,
    c.name,
    sum(p.pcount) as quantity
  from rcte r
  join categories c on c.id = r.ancestor_id
  left join products_per_category p on p.id = r.id
  group by c.id

结果：

| id  | name                 | quantity |
| --- | -------------------- | -------- |
| 1   | ELECTRONICS          | 4        |
| 2   | TELEVISIONS          | 3        |
| 3   | TUBE                 | 1        |
| 4   | LCD                  | 2        |
| 5   | PLASMA               | 0        |
| 6   | PORTABLE ELECTRONICS | 1        |
| 7   | MP3 PLAYERS          | 1        |
| 8   | FLASH                | 0        |
| 9   | CD PLAYERS           | 0        |
| 10  | 2 WAY RADIOS         | 0        |

db-fiddle

如果您想删除空类别 (quantity = 0)，只需将所有 LEFT JOIN 替换为 INNER JOIN。

更新

要获取level，您可以在外部SELECT 中使用子查询：

(select count(*) from rcte r2 where r2.id = c.id) as level

db-fiddle