Ruby on Rails的关系参考节约

Question

我只是想知道什么是节约这类数据

这里的最佳实践年代车型

class User < ActiveRecord::Base 
    has_many :posts, dependent: :destroy 
    has_many :comments, dependent: :destroy 
end 

class Comment < ActiveRecord::Base 
    belongs_to :post 
    belongs_to :user 
end 

class Post < ActiveRecord::Base 
    belongs_to :user 
    has_many :reviews 
end 

目前我保留他们在这之间的关系代码类型

def create 
    @post = Post.find(params[:post_id]) 
    @comment = current_user.reviews.build(comment_param) 
    @comment.post_id = params[:post_id] 
    if @comment.save 
     flash[:success] = 'Successfully created a comment' 
     @post 
    else 
     flash[:error] = 'Error' 
     @post 
    end 
end 

下面是窗体的开始标记。

form_for([@post,@post.comment.build]) do |f| 
    %p 
     = f.label :title, class: 'control-label' 
     = f.text_field :title, class: 'form-control' 
    %p 
     = f.label :message, class: 'control-label' 
     = f.text_area :message, class: 'form-control' 

我如何可以通过快捷键，以便我不会手动分配POST_ID，让轨道解析它

Answer 1

你可以写这样的事情：

@comment = current_user.comments.build(comment_param.merge(:post_id => params[:post_id])) 

或

@post = Post.find(params[:post_id]) 
@comment = @post.comments.build(comment_param.merge(:user => current_user)) 

更新：

或者你可以添加一个scope您Comment型号：

# in comment.rb 
scope :by_user, ->(user) { where(:user_id => user.id) } 

# and then in the controller 
@post = Post.find(params[:post_id]) 
@comment = @post.comments.by_user(current_user).build(comment_param)