搜索帶過濾器的餐廳

Question

嗨，我想寫簡單的搜索引擎。我得到的網址，如：

.../fragments/get_restaurants?city=London&service=1,2 

，我想尋找這樣的：

def get_restaurants 
    regions_find = Region.find(:all, :select => "id", :conditions => ["name LIKE ?", "#{params[:city]}"]).restaurants 
    cities_find = City.find(:all, :select => "id", :conditions => ["name LIKE ?", "%#{params[:city]}%"]).restaurants 
    regions_and_cities = reg.merge!(cit).uniq 
    restaurans_all = regions_and_cities.find(:all, :conditions => ["name LIKE ?", "%#{params[:name]}%" OR TYPE OR KIND]) 

    filtered = filter(restaurans_all) 

    render :partial => "fragments/restaurant", :collection => res 
end 

，但我真的不知道如何使它發揮作用。

這裏是餐廳，市，區模型：

class Region < ActiveRecord::Base 
    has_many :cities, :dependent => :destroy # dependent only with has_ 
    has_many :restaurants 
end 

class Restaurant < ActiveRecord::Base 
    belongs_to :region 
    belongs_to :city 
    has_many :restaurants_types 
    has_many :types, :through => :restaurants_types 
end 

class City < ActiveRecord::Base 
    belongs_to :region 
    has_many :restaurants 
end 

當我有結果，我想通過對其進行過濾：

def filter(table) 
    res = table 
    res &= table.filter('types', params[:type].split(','))   unless params[:type].nil? 
    res &= table.filter('cuisines', params[:cuisine].split(',')) unless params[:cuisine].nil? 
    res &= table.filter('facilities', params[:facility].split(',')) unless params[:facility].nil? 
    res &= table.filter('prices', params[:price].split(','))  unless params[:price].nil? 
    return res 
    end 

與此：

scope :filter, lambda{|type_name, type_id| includes(type_name.to_sym).where(["#{type_name}.id in (?)", type_id]) } 

但它不工作。你能給我一些提示我怎麼能做到這一點？

Answer 1

而不是從頭開始編寫你自己的代碼，我建議使用Searchlogic作爲起點。