2015-10-30 121 views
-1

工作這就是我創建一個會話控制器:會話工作不在服務器上

public function login(){ 
    if(isset($_POST)){ 
     $res = $this->register_model->loginUser(); 
     if($res['result'] === true){ 
      // declare session variables 
      $user = $res['info']; // array containing user information 
      // set session variables 
      $_SESSION['storeId'] = $user->str_id; 
      $_SESSION['sName'] = $user->str_nme; 
      $_SESSION['sId'] = $user->str_identifier; 
      $_SESSION['hash'] = $user->hash; 
      $res['info'] = null; 
     } 
    } else { 
     $res = array('result'=>false,'msg'=>'Login failed. Please try again'); 
    } 
    echo json_encode($res); 
} 

,這是一個控制器,其中i檢索會話

+1

你得到的錯誤是什麼? –

+0

爲什麼混合php會話代碼與你CI會話? – Saty

+0

請提供[最小,完整和可驗證的示例](http://stackoverflow.com/help/mcve) –

回答

1

試試這個

public function login(){ 
    if(isset($_POST)){ 
     $res = $this->register_model->loginUser(); 
     if($res['result'] === true){ 
      // declare session variables 
      $user = $res['info']; // array containing user information 
      // set session variables 

      $this->load->library('session'); 
      $newdata = array(
       'storeId' => $user->str_id, 
       'sName'  => $user->str_nme, 
       'sId' => $user->str_identifier, 
       'hash' => $user->hash, 
       'info' => null, 
       'logged_in' => TRUE, 
      ); 

      $this->session->set_userdata($newdata); 

     } 
    } else { 
     $res = array('result'=>false,'msg'=>'Login failed. Please try again'); 
    } 
    echo json_encode($res); 
} 
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