Rails 嵌套关联并重新编号 [英] Rails nested association and restarted numbering

问题描述

如果我有这样的嵌套资源:

If I have a nested resource like so:

resources :users
  resources :posts
end

和 user has_many posts，是否可以根据 URL 中的父关联让 Rails 开始编号?例如，目前，嵌套资源只是抓取 ID:

and a user has_many posts, it is possible to have Rails start numbering based on the parent association in the URL? For example, currently, nesting resources just grabs the ID:

@user.posts.find(params[:id])

这正确命名了帖子，只允许来自 @user 的帖子......但是，有没有办法使 post_id 是独立的?IE.我希望每个用户的帖子从 1 开始，其中:

This correctly namespaces the posts, only allowing posts from @user... however, is there a way such that the post_id is independent? I.E. I want each user's posts to start at 1, where:

/users/1/posts/1
/users/2/posts/1

实际上是指两个不同的帖子?

Actually refer to two different posts?

推荐答案

这可能需要很多工作，但基本上您可以通过以下步骤完成:

It can be quite a bit of work, but basically you can do it with these steps:

1. 创建迁移以添加新属性来存储特定的用户帖子计数.(我使用了 user_post_id)
2. 覆盖Post 的to_param 方法以使用您刚刚创建的新值.(它必须是一个字符串.)
   • to_param 是 url 和 path 助手使用的方法.

1. Create a migration to add a new attribute to store the specific user-post count. (I used user_post_id)
2. Override Post's to_param method to use the new value you just created. (It has to be a string.)
   • to_param is the method that the url and path helpers use.

更改所有控制器方法以在 user_post_id

Change all your controller methods to find on user_post_id

@user = User.find(params[:user_id])
@post = @user.posts.where(:user_post_id => (params[:id])).first

更改您现在可能无法使用的所有视图

您可以在此处查看源代码:自定义嵌套资源 URL 示例

You can see the source here: Custom Nested Resource URL example

迁移:

class AddUserPostIdToPosts < ActiveRecord::Migration
  def change
    add_column :posts, :user_post_id, :integer
  end
end

post.rb:

class Post < ActiveRecord::Base
  before_save :set_next_user_post_id
  belongs_to :user

  validates :user_post_id, :uniqueness => {:scope => :user_id}

  def to_param
    self.user_post_id.to_s
  end

private
  def set_next_user_post_id
    self.user_post_id ||= get_new_user_post_id
  end

  def get_new_user_post_id
    user = self.user
    max = user.posts.maximum('user_post_id') || 0
    max + 1
  end
end

几个控制器方法post_controller.rb:

A couple controller methods posts_controller.rb:

class PostsController < ApplicationController
  respond_to :html, :xml

  before_filter :find_user

  def index
    @posts = @user.posts.all
    respond_with @posts
  end

  def show
    @post = @user.posts.where(:user_post_id => (params[:id])).first
    respond_with [@user, @post]
  end
  ...
end

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