根据匹配的客户字段（电话、电子邮件、地址）为客户分配 home_id - 重复问题

Question

我有一个客户表，其中每个客户都有一个唯一的id，具体取决于他们下订单时使用的电子邮件。此外，phone、email 和 address 有单独的表列。如果他们的phone、email 或address 与客户表中的另一个“客户”匹配，我正在尝试将客户 ID 分组到同一个 household_id 下。

我遇到的问题是我可以将客户分组在一起并给他们一个household_id，但我很难在过滤中完全删除这些客户分组的重复出现。下面查询中的最后两个 cmets 旨在帮助解释我当前的过滤逻辑并说明它失败的地方。此逻辑适用于成对的客户，但一旦需要将 3 个或更多客户绑定到相同的单个 home_id，就会开始失败。有没有更好的方法来过滤这些结果，或者我是否需要添加一些额外的 CTE，利用 min()/max() 函数和其他类型的分组来在这里添加更多智能？除了 rank() 之外，还有什么其他聪明的窗口函数可以帮助我吗？

with household as (
  select
    c1.id as parent_id,
    c2.id as child_id,
    rank() over (partition by c1.id order by c2.id) as child_number
    -- order by clause is important here to ensure lowest c2.id is always rank 1 (referenced later on in household join onto customer table)
                    
  from customer c1
    left join customer c2 on (c1.phone = c2.phone) or (c1.email = c2.email) or (c1.address = c2.address)
                            
  order by c1.id, child_number
)
                
select
  'H-' || h.parent_id as household_id, -- effectively creates a unique household_id
  h.child_id
                    
from household h
  where h.parent_id < h.child_id or (h.parent_id = h.child_id and h.child_number = 1)
  -- ^this where clause is my attempt at removing the duplicate groupings of customers
  -- it works in the instance when there is a pair of customers tied to a household_id, but when there are 3 or more it starts to fail

查看链接图片以查看家庭 cte 的视图，其中包含 3 个 customer_id 的分组，因为它们具有匹配的电话、电子邮件或地址而连接在一起。突出显示的行是在上述查询的 where 子句中通过我的过滤器的内容
How my query is failing

Answer 1

我使用与 Julius 的 cmets 中讨论的方法类似的方法解决了这个问题：

1. 使用递归 cte grp 将所有相似的客户分组到数组中
2. 在household cte 中使用select distinct on 和cardinality() 函数将递归grp 缩减为仅包含所有家庭成员的数组
3. 扫描household 表并通过取消嵌套linked_customers 数组并根据数组中的最小ID 将其所有元素绑定到单个household_id 来删除所有重复项

这似乎工作得很好，但我确信这个查询可以进一步简化，我很乐意欢迎任何反馈！

with recursive grp as (
    select
        c1.id,
        c1.email,
        c1.phone,
        c1.address,
        array[c1.id] as linked_customers -- initializes an array based of the id of every customer
    from customer c1
    
    union all
    
    select
        c2.id,
        c2.email,
        c2.phone,
        c2.address,
        c2.id || linked_customers
    from grp g
        join customer c2 on (g.email = c2.email or g.phone = c2.phone or g.address = c2.address)
    where c2.id <> all(linked_customers) -- ensures the same customer id that already was used in the array initialization is not being looked at again
), -- creates several similar groups as well as intermediary groups of customers

household as (
    select
        distinct on (g.id) g.linked_customers
    from grp g
    order by g.id, cardinality(linked_customers) desc
) -- extracts largest array for each customer_id (still duplicate groupings here), but only the max length arrays are being pulled out (all household members)

select
    distinct on (p.parent_id) 'H-' || parent_id as household_id,
    unnest(p.linked_customers) as child_id
from (
    select
        min(parent_id) as parent_id, -- pulls out the minimum id of each linked customers group which will remove the creation of multiple household_ids for the same customer groups in the select clause above
        h.linked_customers
    from household h, unnest(h.linked_customers) parent_id
    group by h.linked_customers
) p
order by parent_id