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我有两张简单的桌子。首先是品牌，每个品牌都可能有一个顶级品牌。例如 Elseve 有一个顶级品牌是 Loreal Paris。但雅芳没有顶级品牌。我有一个简单的产品表。

这是sqlfiddle

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，产品 1 文档返回空结果。但我需要在文档中至少获得雅芳。

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

如何解决这个问题呢 ？

感谢 Voa Tsun。我稍微更新了查询。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;
4

1 回答 1

1

正如我从您的帖子中得到的那样，基本思想是在“parent_id”中加入 id 而不是针对 null，而是“至少针对 id”。像这儿：

   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
于 2017-05-31T13:08:55.867 回答