问题描述
我有 3 张桌子
items (columns are: name , type)
history(columns are: date, username, item_id)
user(username, password)
当用户说ABC"登录并创建新项目时,将使用以下 after_create 过滤器创建历史记录.如何通过此过滤器将此用户名ABC"分配给历史表中的用户名字段.
When a user say "ABC" logs in and creates a new item, a history record gets created with the following after_create filter.How to assign this username ‘ABC’ to the username field in history table through this filter.
class Item < ActiveRecord::Base
has_many :histories
after_create :update_history
def update_history
histories.create(:date=>Time.now, username=> ?)
end
end
我在 session_controller 中的登录方法
My login method in session_controller
def login
if request.post?
user=User.authenticate(params[:username])
if user
session[:user_id] =user.id
redirect_to( :action=>'home')
flash[:message] = "Successfully logged in "
else
flash[:notice] = "Incorrect user/password combination"
redirect_to(:action=>"login")
end
end
end
我没有使用任何身份验证插件.如果可能的话,如果有人能告诉我如何在不使用插件(如 userstamp 等)的情况下实现这一点,我将不胜感激.
I am not using any authentication plugin. I would appreciate if someone could tell me how to achieve this without using plugin(like userstamp etc.) if possible.
推荐答案
Rails 5
声明一个模块
module Current
thread_mattr_accessor :user
end
分配当前用户
class ApplicationController < ActionController::Base
around_action :set_current_user
def set_current_user
Current.user = current_user
yield
ensure
# to address the thread variable leak issues in Puma/Thin webserver
Current.user = nil
end
end
现在您可以将当前用户称为 Current.user
Now you can refer to the current user as Current.user
在模型中访问 current_user
并不常见.话虽如此,这里有一个解决方案:
It is not a common practice to access the current_user
within a model. That being said, here is a solution:
class User < ActiveRecord::Base
def self.current
Thread.current[:current_user]
end
def self.current=(usr)
Thread.current[:current_user] = usr
end
end
在 ApplicationController
的 around_filter
中设置 current_user
属性.
Set the current_user
attribute in a around_filter
of ApplicationController
.
class ApplicationController < ActionController::Base
around_filter :set_current_user
def set_current_user
User.current = User.find_by_id(session[:user_id])
yield
ensure
# to address the thread variable leak issues in Puma/Thin webserver
User.current = nil
end
end
认证成功后设置current_user
:
def login
if User.current=User.authenticate(params[:username], params[:password])
session[:user_id] = User.current.id
flash[:message] = "Successfully logged in "
redirect_to( :action=>'home')
else
flash[:notice] = "Incorrect user/password combination"
redirect_to(:action=>"login")
end
end
最后参考Item
的update_history
中的current_user
.
class Item < ActiveRecord::Base
has_many :histories
after_create :update_history
def update_history
histories.create(:date=>Time.now, :username=> User.current.username)
end
end
这篇关于访问模型中的 current_user的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持!