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Php 模型中的Codeigniter静态变量_Php_Codeigniter_Logging_Model_Static - Fatal编程技术网

Php 模型中的Codeigniter静态变量

Php 模型中的Codeigniter静态变量,php,codeigniter,logging,model,static,Php,Codeigniter,Logging,Model,Static,一旦用户登录-我希望我的所有模型都知道用户的id。即使以后调用它们 我曾想过使用静态变量,但它似乎不起作用 class Base_model extends CI_Model { static protected $user_id; } class Log_in_model extends Base_model { public function log_in(){ self::$user_id = 69; } } class A_model exte

一旦用户登录-我希望我的所有模型都知道用户的id。即使以后调用它们

我曾想过使用静态变量,但它似乎不起作用

class Base_model extends CI_Model {
    static protected $user_id;
}

class Log_in_model extends Base_model {
    public function log_in(){
        self::$user_id = 69;
    }
}

class A_model extends Base_model {
    public function do_A(){
        echo self::$user_id;
    }
}

class B_model extends Base_model {
    public function do_B(){
        echo self::$user_id;
    }
}
初始化会话

$this->load->library('session');
用户登录后,将userdata保存在会话userdata中

$newdata = array(
               'username'  => 'USERNAME',
               'email'     => 'EMAIL',
               'user_id'   => 'USERID',
               'logged_in' => TRUE
           );

$this->session->set_userdata('userdetails',$newdata);   //setting data in session with a name userdetails 
获取会话用户数据

 print_r($this->session->userdata('userdetails'));  //get userdetails from session
要从会话中销毁userdetails,请使用

$this->session->unset_userdata('userdetails'); 
如果你想了解更多有关会话的信息,请阅读此


为什么你不使用会话…?仅仅基于会话数据让用户进入系统不是有安全风险吗?不。您可以使用$config['sess\u encrypt\u cookie']=TRUE来保护它$config['sess_use_database']=TRUE@马说了这一切…: