问最快的方式，通过匹配，有许多通过关联？

EN

class Product < ActiveRecord::Base
  has_many :product_taggings
  has_many :product_tags, :through => :product_taggings
end

class ProductTags < ActiveRecord::Base
  has_many :product_taggings
  has_many :products, :through => :product_taggings
end

class ProductTaggings < ActiveRecord::Base
  belongs_to :product
  belongs_to :product_tags
end

SELECT
  `product_id`
  ,COUNT(*) AS relevance
FROM
  `product_taggings` AS ptj
LEFT JOIN
  `products` AS p
    ON p.`id` = ptj.`product_id`
LEFT JOIN
  `product_tags` AS pt
    ON pt.`id` = ptj.`product_tag_id`
WHERE
  pt.`name` IN ('Tag 1', 'Tag 2')
GROUP BY
  `product_id`

Product 1 -> Tag 3
Product 2 -> Tag 1, Tag 2
Product 3 -> Tag 1, Tag 3

product_id | relevance
----------------------
         2 |         2
         3 |         1

* Product 1 is not included since there were no matches.
  Given that the user is performing a filtered search,
  this behavior is probably fine.  There's a way to get
  Product 1 into the results with 0 relevance if
  necessary.

SELECT *
FROM
  `products` AS p
  ,(SELECT
      `product_id`
      ,COUNT(*) AS relevance
    FROM
      `product_taggings` AS ptj
    LEFT JOIN
      `products` AS p
        ON p.`id` = ptj.`product_id`
    LEFT JOIN
      `product_tags` AS pt
        ON pt.`id` = ptj.`product_tag_id`
    WHERE
      pt.`name` IN ('Tag 1', 'Tag 2')
    GROUP BY `product_id`
  ) AS r
WHERE
  p.`id` = r.`product_id`
ORDER BY
  r.`relevance` DESC

def Product.find_by_tag_list(tag_list)
  tag_list_sql = "'" + tag_list.join("','") + "'"
  Product.find_by_sql("SELECT * FROM products, (SELECT taggable_id, COUNT(*) AS relevance FROM taggings LEFT JOIN tags ON tags.id = taggings.tag_id WHERE tags.name IN (" + tag_list_sql + ") GROUP BY taggable_id) AS r WHERE products.id = r.taggable_id ORDER BY r.relevance DESC;")
end

Product.find_by_tag_list(my_product.tag_list)