Question

我有两个简单的表格。首先是品牌，每个品牌都可能拥有顶级品牌。例如，Elseve拥有顶级品牌Loreal Paris。但雅芳没有顶级品牌。我有一个简单的产品表。

这是品牌表。

 id |     name     | parent_id | depth
----+--------------+-----------+-------
  3 | Loreal Paris |     null  |     0
  1 | Avon         |     null  |     1
  2 | Elseve       |      3    |     1
(3 rows)

这是产品表

 id | brand_id |   name
----+----------+-----------
  1 |        1 | Product 1
  2 |        2 | Product 2
(2 rows)

当我尝试获取tsvectors时，Product 1文档返回null结果。但我需要在文件中至少获得雅芳。

 product_id | product_name |                     document
------------+--------------+--------------------------------------------------
          1 | Product 1    |
          2 | Product 2    | '2':2 'elsev':3 'loreal':4 'paris':5 'product':1
(2 rows)

如何解决这个问题？

感谢Voa Tsun。我稍微更新了一下查询。我不再需要分组了。

select
  products.id as product_id,
  products.name as product_name,
  to_tsvector(products.name) ||
  to_tsvector(brands.name) ||
  to_tsvector(top_brands.name)
    as document
from products
  JOIN brands on brands.id = products.brand_id
  LEFT JOIN brands as top_brands on coalesce(brands.parent_id,brands.id) = top_brands.id;

Answer 1

基本的想法是在＆＃34; parent_id＆＃34;中加入id而不是null，但是＃34;至少对于id＆＃34;，正如我从你的帖子中得到的那样。像这里：

   select
      products.id as product_id,
      products.name as product_name,
      to_tsvector(coalesce(products.name,brands.name)) ||
      to_tsvector(brands.name) ||
      to_tsvector(coalesce(string_agg(top_brands.name, ' ')))
        as document
    from products
      JOIN brands on brands.id = products.brand_id
      LEFT JOIN brands as top_brands on coalesce(brands.parent_id,brands.id) = 
     top_brands.id
    GROUP BY products.id,brands.id;

    product_id  product_name    document
1   1   Product 1   '1':2'avon':3,4'product':1
2   2   Product 2   '2':2'elsev':3'loreal':4'pari':5'product':1

Postgres tsvector与关系表

1 个答案: