Question

我想创建一个Current_User方法，但我不想使用gem或类似方法。我将如何在Rails 4.1.2

中做到这一点

Questions_Controller我想要current_user方法。

class QuestionsController < ApplicationController
before_filter :auth, only: [:create, :your_questions, :edit, :update]

  # def index
  #     @question = Question.new
  #   @questions = Question.unsolved(params)
  # end

  @questions = current_user.your_questions(params[:page])

  def your_questions(page)
  questions.paginate(page: page, order: 'created_at DESC', per_page: 3)
  end


  def self.unsolved(params)
    order('created_at DESC').where(solved: false).paginate(page: params[:page],per_page: 3)
  end

  def create
    @question = current_user.questions.build(params[:question])
    if @question.save
        flash[:success] = 'Your question has been posted!'
        redirect_to @question
    else
      @questions = Question.unsolved(params)
        render 'index'
    end
  end

  def new
       @question = Question.new
  end

  def show
    # raise FOO
    puts params
    @question = Question.find(params[:id])
    @answer = Answer.new
  end

  def your_questions
    @questions = current_user.your_questions(params[:page])
    # current_user.your_questions(params[:id])
  end

  def edit
    @question = current_user.questions.find(params[:id])
  end

  def update
    @question = current_user.questions.find(params[:id])

    if @question.update_attributes(params[:question])
      flash[:success] = 'Your question has been updated!'
      redirect_to @question
    else
      render 'edit'
    end
  end

  def search
    @questions = Question.search(params)
  end
end

我的用户模型

    class User < ActiveRecord::Base
    has_many :questions
    has_many :answers
  # attr_accessible :username, :password, :password_confirmation

  has_secure_password

  # validates :username, presence: true, uniqueness: { case_sensitive: false },
  #                                          length: { in: 4..12 },
  #                                          format: { with: /A[a-z][a-z0-9]*z/, message: 'can only contain lowercase letters and numbers' }
    validates :password, length: { in: 4..8 }
    validates :password_confirmation, length: { in: 4..8 }

    def your_questions(page)
    questions.paginate(page: page, order: 'created_at DESC', per_page: 3)
  end
end

我的应用程序控制器

    class ApplicationController < ActionController::Base
  protect_from_forgery with: :exception
  # protect_from_forgery

  helper_method [:current_user, :logged_in?]

  protected

  private 

    def login(user)
        session[:user_id] = user.id
    end

    def current_user
        current_user ||= User.find(session[:user_id]) if session[:user_id]
    end

    def logged_in?
        !current_user.nil?
    end

    def auth
      redirect_to login_url, alert: 'You must login to access that page' unless logged_in?
    end
end

如果有更多文件需要我添加到问题中，请在rails开发者评论我是一个新手ruby：）

Answer 1

class ApplicationController < ActionController::Base

 helper_method :current_user 

 private

 def current_user
  @current_user ||= User.find(session[:user_id]) if session[:user_id]
 end

end

Answer 2

我认为您的问题不在控制器中，而在于您的助手。

正在调用寻找局部变量current_user的方法而且没有一个方法。你需要像这样实例化这些：

class ApplicationController < ActionController::Base
  protect_from_forgery with: :exception

  helper_method [:current_user, :logged_in?]


  private 

    def login(user)
      session[:user_id] = user.id
    end

    def current_user
       @current_user ||= User.find(session[:user_id]) if session[:user_id]
    end

    def logged_in?
      !current_user.nil?
    end

    def auth
      redirect_to login_url, alert: 'You must login to access that page' unless logged_in?
    end
end

喜欢@RafalG。声明请参阅@前面的current_user。这将创建一个实例变量来跟踪，而不是引用缺少的局部变量。

另请注意，您当前的current_user方法将始终运行User.find，因为本范围内的局部变量current_user始终为nil，因此您需要进行此操作实例的一部分。

<强>更新

我将保留以上内容以进行启发，因为您仍应创建实例。我认为这是真正的问题

class QuestionsController < ApplicationController
  before_filter :auth, only: [:create, :your_questions, :edit, :update]

  # def index
  #   @question = Question.new
  #   @questions = Question.unsolved(params)
  # end

  #vvvv This Line is out of a scope and will raise errors vvv#
  @questions = current_user.your_questions(params[:page])

  def your_questions(page)
    questions.paginate(page: page, order: 'created_at DESC', per_page: 3)
  end
  ....
end

如果你想这样做，你会在before_filter回调中声明它，因为现在Rails不知道如何恰当地处理这个语句，并且在一个它无法访问的方法之外你的任何助手。

如何在不使用设计的情况下创建current_user方法

2 个答案: