Question

鉴于以下结构

CREATE TABLE products (
    id integer NOT NULL,
    subcategory_id integer,
    stack_id integer,
)

CREATE TABLE subcategories (
  id integer NOT NULL,
  name character varying(255)
)

其中products.stack_id是回归产品的自我参照关系。

我基本上试图计算子类别加入产品

products.subcategory_id = subcategories.id

但是每个不同的堆栈组将计数限制为一次。

样本子类别表

id    name
1     subcategory_1
2     subcategory_2
3     subcategory_3

样本产品表

id    subcategory_id    stack_id    
1     1                 NULL        
2     1                 1           
3     2                 1           
4     3                 1           
5     2                 NULL        
6     2                 5           
7     2                 5           
8     2                 NULL        
9     3                 8           
10    3                 8

示例所需的输出

id    name             total 
1     subcategory_1    1     (row 1)
2     subcategory_2    3     (row 1 + row 5 + row 8)
3     subcategory_3    2     (row 1 + 8)

输出说明

子类别ID 1
如果我与产品进行简单的连接，我会得到产品（1,2）。我只想要不同父对象的数量（stack_id为null），因此1个计数和2个已经计数的引用1因此不会增加计数。

子类别ID 2
加入将是（3,5,6,7,8）。 3的stack_id是1所以它计数1.产品5,6和7参考5，因此计数1.产品8计数1.

子类别3
加入是（4,9,10）。 4参考文献1，9和10都参考了8.

更新

删除了额外可能令人困惑的列，添加了样本数据和输出

Answer 1

如果引用的最大深度是一个级别，那么这个简单的查询就完成了这项工作：

select subcategory_id, name, count(*)
from (
    select distinct subcategory_id, coalesce(stack_id, id) stack_id
    from products
    ) sub
join subcategories s on s.id = sub.subcategory_id
group by 1, 2
order by 1, 2;

 subcategory_id |     name      | count 
----------------+---------------+-------
              1 | subcategory_1 |     1
              2 | subcategory_2 |     3
              3 | subcategory_3 |     2
(3 rows)

此递归查询也适用于深度超过一个级别的引用：

with recursive pr(id, subcategory_id, stack_id, stack) as (
    select id, subcategory_id, stack_id, array[id]
    from products
union
    select pr.id, pr.subcategory_id, products.stack_id, pr.stack_id || pr.stack
    from pr
    join products on pr.stack_id = products.id
    )
select distinct on (id) id, subcategory_id, stack
from pr
order by id, array_length(stack, 1) desc

 id | subcategory_id | stack  
----+----------------+--------
  1 |              1 | {1}
  2 |              1 | {1,2}
  3 |              2 | {1,3}
  4 |              3 | {1,4}
  5 |              2 | {5}
  6 |              2 | {5,6}
  7 |              2 | {5,7}
  8 |              2 | {8}
  9 |              3 | {8,9}
 10 |              3 | {8,10}
(10 rows)

使用上述数据集加入子类别：

select subcategory_id, name, count(*)
from (
    select distinct subcategory_id, stack[1]
    from (
        with recursive pr(id, subcategory_id, stack_id, stack) as (
            select id, subcategory_id, stack_id, array[id]
            from products
        union
            select pr.id, pr.subcategory_id, products.stack_id, pr.stack_id || pr.stack
            from pr
            join products on pr.stack_id = products.id
            )
        select distinct on (id) id, subcategory_id, stack
        from pr
        order by id, array_length(stack, 1) desc
        ) sub
    ) sub
join subcategories s on s.id = sub.subcategory_id
group by 1, 2
order by 1, 2

 subcategory_id |     name      | count 
----------------+---------------+-------
              1 | subcategory_1 |     1
              2 | subcategory_2 |     3
              3 | subcategory_3 |     2
(3 rows)

Postgres计数与自引用连接条件

1 个答案: