轨道与搜索关联

Question

>我对与搜索的关联有问题

class Manufacturer < ApplicationRecord
  has_many :translations, class_name: 'ManufacturerTranslation', dependent: :destroy
def self.search(query)
    q = query.to_s.parameterize.gsub('-',' ').split
    joins(:translations).where("lower(name) LIKE ?", "%#{q}%")
end
and 

 class ManufacturerTranslation < ApplicationRecord
  belongs_to :manufacturer
end

因此，当我尝试对其进行搜索并调用翻译时

Manufacturer.search('fra').last.translations

它只给了我名称包含fra的翻译，而不是所有的翻译

所以我总共为这个制造商提供了 6 个翻译

但是搜索后只得到2

数据库架构

 create_table "manufacturer_translations" do |t|
    t.integer "manufacturer_id"
    t.string "locale"
    t.string "name"
    t.string "image_source_url"
    t.datetime "created_at", null: false
    t.datetime "updated_at", null: false
    t.index ["manufacturer_id"], name: "index_manufacturer_translations_on_manufacturer_id"
    t.index ["name"], name: "index_manufacturer_translations_on_name"
  end
  create_table "manufacturers", do |t|
    t.datetime "created_at", null: false
    t.datetime "updated_at", null: false
    t.boolean "deleted", default: false
  end

Answer 1

在拆分字符串时，您必须遍历所有实体才能搜索它们。而且您还必须将结果缩小。尝试以下代码：

def self.search(query)
    q          = query.to_s.parameterize.gsub('-',' ').split
    conditions = ''
    q.each do |qu|
      conditions = "#{conditions} OR " if conditions.present?
      conditions = "#{condidtions} lower(manufacturers.name) LIKE %#{qu.downcase}%"
    end
    joins(:translations).where(conditions)
end

Answer 2

假设你遵循 Rails 约定，我建议你替换它

joins(:translations).where("lower(name) LIKE ?", "%#{q}%")

跟

joins(:translations).where("lower(manufacturers.name) LIKE ?", "%#{q}%")

可能需要玩弄引号