Rails 控制器中实例变量值的初始计算错误

Question

我希望构建一个基本应用程序，该应用程序可以根据用户对给定午餐的愉快程度进行评分。

在我看来，投票按钮是这样的：

<%= form_for :lunch, url: upvote_lunch_lunch_path(params[:id]), method: :put, :html => {:class => 'form-inline'} do |f| %>
  <%= f.hidden_field :liked %>
  <%= f.hidden_field :id %>
  <%= f.submit "Like Lunch", class: "btn btn-large btn-success" %>
<% end %>

<%= form_for :lunch, url: downvote_lunch_lunch_path(params[:id]), method: :put, :html => {:class => 'form-inline'} do |f| %>
  <%= f.hidden_field :disliked %>
  <%= f.hidden_field :id %>
  <%= f.submit "Dislike Lunch", class: "btn btn-large btn-danger" %>
<% end %>

显示值（部分）如下所示：

<%= number_to_percentage(@enjoy_score * 100, precision: 0, format: "%n") %>

最后，在我的 "lunches" 控制器中，我有以下内容：

def show
    @lunch = Lunch.find(params[:id])
    @provider = @lunch.provider
    @total_enjoy_votes = (@lunch.liked + @lunch.disliked)
    @enjoy_score = (@lunch.liked.to_f / @total_enjoy_votes)

  end

  def downvote_lunch
    @lunch = Lunch.find(params[:id])
    @lunch.increment!(:disliked)
    redirect_to @lunch
  end

  def upvote_lunch
    @lunch = Lunch.find(params[:id])
    @lunch.increment!(:liked)
    redirect_to @lunch
  end

一切都按预期工作，只要数据库已经有针对特定午餐 ID 的喜欢和不喜欢的值。例如，如果您是第一个尝试回答此问题的人（假设刚创建午餐），应用程序将出错并显示消息“undefined method +' for nil:NilClass" at this line:@total_enjoy_votes = (@ lunch.liked + @lunch.disliked)`

两个问题：

为什么如果我打开数据库（使用 sqlitebrowser）和 "seed" 喜欢 = 1 和不喜欢 = 1 的一行数据，这些方法将按预期工作，但如果我不这样做，它出错了吗？我应该如何处理 "initialize" @lunch.liked 和 @lunch.disliked 以便初始用户不会出错？
（奖励积分）如何让控制器代码保持干燥，这样我就不必输入 @lunch = Lunch.find(params[:id]) 在每个方法的开头？

非常感谢。如果这是一个非常简单的问题，我深表歉意。

Answer 1

1.

@total_enjoy_votes = (@lunch.liked + @lunch.disliked) 错误，因为 @lunch.liked 是 nil，因为它从未设置为任何东西，以及 @lunch.disliked

为避免此错误，您应该检查是否存在喜欢和不喜欢的内容。

liked = @lunch.liked ? @lunch.liked : 0
disliked = @lunch.disliked ? @lunch.disliked : 0
@total_enjoy_votes = (liked + disliked)
@enjoy_score = (liked.to_f / @total_enjoy_votes)

2.

before_filter :find_lunch!, only: [ :update, :destroy ] # list of actions where to perform the method.
private  
 def find_lunch!
   @lunch = Lunch.find(params[:id])
 end

编辑

对行的解释：@lunch.liked ? @lunch.liked : 0

经常派上用场的是ternary operator

语法

boolean_expression ? true_expression : false_expression

例子

grade = 88
status = grade >= 70 ? "pass" : "fail"
#=> pass

是一样的，好像我写的是这样的：

if @lunch.liked.nil?
  liked = 0
else
  liked = @lunch.liked
end

Answer 2

第一种方法：-

def show
  @lunch = Lunch.find(params[:id])
  @provider = @lunch.provider
  liked = ((@lunch.liked.present? and @lunch.liked >= 0) ? @lunch.liked : 0)
  @total_enjoy_votes = liked + ((@lunch.disliked.present? and @lunch.disliked >= 0) ? @lunch.disliked : 0)
  @enjoy_score =  (@total_enjoy_votes > 0 and liked > 0) ? (@lunch.liked.to_f / @total_enjoy_votes) : 0
end

第二种方法：-

在 Lunch model 中设置回调为

class Lunch <  < ActiveRecord::Base
  before_save :initialize_liked_disliked

  private
  def initialize_liked_disliked
    if self.new_record?
      self.liked = 0
      self.disliked = 0
    end
  end
end

它将用 0 初始化两个字段，然后在控制器中进行更改：

def show
  @lunch = Lunch.find(params[:id])
  @provider = @lunch.provider
  @total_enjoy_votes = @lunch.liked + @lunch.disliked
  @enjoy_score =  (@total_enjoy_votes > 0) ? (@lunch.liked.to_f / @total_enjoy_votes) : 0
end

Rails 控制器中实例变量值的初始计算错误

Error with initial calculation of instance variable values in Rails controller

methods

ruby-on-rails

erb

ruby-on-rails-4

编辑