Question

我是ruby on rails的新手，我想为每个已登录的用户创建一个会话。但我得到了那个错误。在我的控制器中，我有：

class SessionsController < ApplicationController
def new
end

def create
user =User.find_by pseudo :params[:pseudo]
if user
  session[:user_id]=user.id
  flash[:notice]= "you are signed in! "
  redirect root_url
else
  flash.now[:alert]= "wrong password/pseudo"
  render 'new'
end
end

def destroy
session[:user_id]=nil
flash[:notice]='You are now signed out'
redirect root_url
end
end

在我的new.html.erb中：

<h1>Sessions#new</h1>
<%= form_tag sessions_path do  %>
<div>
<%=label_tag 'pseudo', 'pseudo' %>
<%= text_field_tag 'pseudo' %>
</div>
<div>
  <%= label_tag 'password', 'password' %>
  <%= password_field_tag 'password' %>
</div>
  <%= submit_tag 'log in' %>


<% end %>

在我的索引中我有：

<h1>Users#index</h1>
<% if current_user.present? %>
Welcome <%= current_user.pseudo %> | <%= link_to 'Sign out' , logout_path %>

<% else %>
<%= link_to 'Register', new_user_url %><br/>
<%= link_to 'SignIn', login_path %>
<% end %>

在application_controller.rb中：

 class ApplicationController < ActionController::Base
 # Prevent CSRF attacks by raising an exception.
 # For APIs, you may want to use :null_session instead.
  protect_from_forgery with: :exception

  helper_method :current_user
  private
  def current_user
  User.find_by id: session[:user_id] if session[:user_id]
  end
  end

并且该错误与＆＃34; user = User.find_by pseudo：params [：pseudo]＆＃34;相关。我不明白为什么

错误：没有将Symbol隐式转换为Integer

0 个答案: