所以,我正在构建一个教育资源搜索引擎Rails应用程序。这个应用程序将以块式布局显示结果(例如微积分视频),旨在利用Isotope(用于冷却过滤/排序转换等)。
除了查询我们的数据库以获取适当的资源外,我们还通过他们的自定义搜索引擎API查询Google。最重要的是,我们最终会想要将一些广告块放入网格视图中。
所以我的问题是,如何将查询数据库时返回的@resources与Google查询的JSON结果以及最终的广告相结合?我想这样做,以使结果视图(search.html.erb)尽可能干净。此外,我希望能够对所有结果进行排序/过滤。也就是说,我希望能够将ActiveRecord结果与Google查询结果合并。这也可以让我做以下事情:
(Boxx是我想到的通用类)
<% @boxxes.each do |boxx| %>
<div class=<%= boxx.type %>>
<h2><%= boxx.title %></h2>
<h3><%= boxx.description %></h3>
...
...
...
</div>
<% end %>
我的资源控制器在下面。基本上,我想将@resource与Google查询的结果合并为一个具有公共接口的可枚举,并且我可以根据用户指定的排序类型进行排序。
最好的方法是什么?我应该在控制器的底部创建一个Boxx类,并能够使用资源,谷歌JSON或广告来初始化它吗?然后我可以保留一个类型变量,然后能够将它们一起排序吗?
这是我的资源控制器
require 'will_paginate/array'
class ResourcesController < ApplicationController
def index
@resources = Resource.all
end
def create
# Usability concern here... need to make sure that they are redirected back here once they log in or something
if current_user == nil
flash[:alert] = "You must log in to submit a resource!"
redirect_to resources_path
return
else
params[:resource][:user_id] = current_user.id
end
# Check to see if this resource unique
params[:resource][:link] = Post::URI.clean(params[:resource][:link])
if unique_link?(params[:resource][:link])
@resource = Resource.new(params[:resource])
@resource[:youtubeID] = self.isYoutube(@resource[:link])
@resource.save
else
flash[:alert] = "This resource has already been added!"
end
redirect_to resources_path
end
def vote
value = params[:type] == "up" ? 1 : -1
@resource = Resource.find(params[:id])
@resource.add_or_update_evaluation(:votes, value, current_user)
respond_to do |format|
format.html { redirect_to :back, notice: "Thank you for voting" }
format.json { render :status=>200, :json=>{:success=>true}}
end
end
def isYoutube(youtube_url)
regex = %r{http://www.youtube.com}
if youtube_url[regex]
youtube_url[/^.*((v\/)|(embed\/)|(watch\?))\??v?=?([^\&\?]*).*/]
youtube_id = $5
thumbnail_Link = "http://img.youtube.com/vi/#{youtube_id}/1.jpg"
else
thumbnail_Link = nil
end
thumbnail_Link
end
def unique_link?(url)
Resource.find_by_link(url) == nil
end
def search
@resource = Resource.full_search(params[:q])
# raise params.to_s
@resource = @resource.reject!{|r| !r.media_type.eql? params[:filter][0][:media_type].downcase } if params[:filter] && !params[:filter][0][:media_type].blank?
if params[:filter]
case params[:filter][0][:sort].downcase
when 'newest'
then @resource = @resource.sort_by{|r| r.created_at}
when 'votes'
then @resource = @resource.sort_by!{|r| r.reputation_for(:votes).to_i}.reverse
else
end
end
@resource = @resource.paginate(:page => (params[:page] || 1), :per_page => 15)
end
def google(q, filter)
# Authenticating into Google's API
client = Google::APIClient.new(:key => 'secret', :authorization => nil)
# Discover the Custom Search API
search = client.discovered_api('customsearch')
# Search Google CSE
response = client.execute(
:api_method => search.cse.list,
:parameters => {
'q' => "#{q} #{filter}",
'key' => 'secret',
'cx' => 'secret'
}
)
# Decode the results
results = ActiveSupport::JSON.decode(response.body, {:symbolize_names => true})
# Return an empty array if Google CSE limit has been met.
results["items"] == nil ? [] : results["items"]
end
def make_boxxes(resources, google_results, ads)
end
end
编辑#1:等等,可以创建一个GoogleResult类,然后执行
@items = @resources | google_results
因为我可以让GoogleResult遵循与Resources相同的界面。但那我该如何排序呢?嗯...
答案 0 :(得分:0)
通过阅读“在Ruby on Rails中设计助手”
,我找到了我所需要的东西http://techspry.com/ruby_and_rails/designing-helpers-in-ruby-on-rails/