如何对一张照片进行投票并自动对另一张照片进行投票？

Question

我有一个显示 2 张随机照片的竞赛模型的显示视图：

<% if @contest.photos.size < 2 %>
    <p>not enough photos in this contest</p>
<% else %>
    <%= link_to (image_tag @random_photo1.image.url(:thumb)), upvote_path(id: @contest.id, photo_id: @random_photo1.id) %>
    <%= @random_photo1.score %>
    <%= link_to (image_tag @random_photo2.image.url(:thumb)), "#" %>
    <%= @random_photo2.score %>
<% end %>

这是我的表演动作：

  def show
    @contest = Contest.find(params[:id])
    @user = current_user

    random_two_photos = @contest.photos.limit(2).order("RANDOM()")
    @random_photo1 = random_two_photos[0]
    @random_photo2 = random_two_photos[1]
  end

这是我的 2 个动作的路线：

  get 'contest/:id/photo/:photo_id' => 'contests#enter_contest', as: :enter_contest
  get 'contest/:id/photo/:photo_id' => 'contests#upvote', as: :upvote

如何编写 "upvote" 控制器动作，使我点击的照片获得 100 分，而另一张照片获得 -100 分？我正在使用 postgresql。

Answer 1

我会使用其他路线：

get 'contests/:id/start' => 'contests#start', as: start_contest
put 'contests/:id/vote' => 'contests#vote', as: vote

然后，在contests_controller/start动作中

def start
  @contest = Contest.find(params[:id])
  random_photos = @contest.photos.limit(2).order("RANDOM()")
  @photo1 = random_photos.first
  @photo2 = random_photos.last
end

在contests_controller/start行动中

def vote
  @contest = Contest.find(params[:id])
  @photo1 = Photo.find(params[:up_photo_id])
  @photo2 = Photo.find(params[:down_photo_id])
  @photo1.upvote # you need to put this method in Photo model
  @photo2.downvote # you need to put this method in the Photo model
end

在contests/start视图中：

<%= link_to image_tag(@photo1.image.url(:thumb)), vote_path(id: @contest.id, up_photo_id: @photo1.id, down_photo_id: @photo2.id) %>

和...

<%= link_to image_tag(@photo2.image.url(:thumb)), vote_path(id: @contest.id, up_photo_id: @photo2.id, down_photo_id: @photo1.id) %>

Remember: you need to pass the option method: :put to link_to helper in order to perform a PUT instead a GET.

改进

您可以将代码放在 vote 动作中 Transaction. It's not a clean solution to put transactions in the controller layer. Maybe you can use the Service pattern
您可以将 "links to" 方法移动到视图助手。类似于：vote_link(contest, up_photo, down_photo)避免代码重复。
您可以将此：photos.limit(2).order("RANDOM()") 移动到模型中的方法。类似于：@contest.random_photos(2)

The code I wrote was not tested. But you get the idea...

更新以使用单个表演动作

您需要定义一条路线

resources :contests, only: [:show]

然后在contests_controller/show动作

def show
  @contest = Contest.find(params[:id])

  if params.has_key? :up_photo_id && params.has_key? :down_photo_id
    @photo1 = Photo.find(params[:up_photo_id])
    @photo2 = Photo.find(params[:down_photo_id])
    @photo1.upvote # you need to put this method in Photo model
    @photo2.downvote # you need to put this method in the Photo model
  else 
    random_photos = @contest.photos.limit(2).order("RANDOM()")
    @photo1 = random_photos.first
    @photo2 = random_photos.last
  end
end

在show.html.erb

<%= link_to image_tag(@photo1.image.url(:thumb)), contest_path(id: @contest.id, up_photo_id: @photo1.id, down_photo_id: @photo2.id) %>

和...

<%= link_to image_tag(@photo2.image.url(:thumb)), contest_path(id: @contest.id, up_photo_id: @photo2.id, down_photo_id: @photo1.id) %>

更新：获取不同的照片

class Contest < ActiveRecord::Base

  def random_photos
    self.photos.order("RANDOM()")
  end

  def two_different_photos
    photo1 = random_photos.first
    raise 'no photos' unless photo1 # raise an exception or something
    photo2 = random_photos.where.not(id: photo1.id).first
    raise 'only one photo' unless photo2 # raise an exception or something
    [photo1, photo2]
  end
end

然后你可以在你的控制器中使用@contest.two_different_photos...

如何对一张照片进行投票并自动对另一张照片进行投票？

How to upvote a photo and automatically downvote the other photo?

postgresql

ruby-on-rails

rails-activerecord