2013-11-15 53 views
0

在下面的codeigniter代碼中我已經放置了控制器,如果我輸入無效的用戶名或passwors它顯示非活動用戶請聯繫管理員,但我想顯示invalisd用戶名或密碼。請幫我解決問題。錯誤if if條件

控制器:

function inactive() 
    { 
    echo"<script>alert('In active user Please contact the administrator');</script>"; 
    $this->load->view('login_form'); 
    } 
    function invalid() 
    { 
    echo"<script>alert('Invalid username or password');</script>"; 
    $this->load->view('login_form'); 
    } 
    function validate_credentials() 
    {  
     $this->load->model('membership_model'); 
     $query = $this->membership_model->validate(); 

     if($query) // if the user's credentials validated... 
     { 
      $data = array(
       'username' => $this->input->post('username'), 
       'is_logged_in' => true 
      ); 
      if($query->num_rows()>0){ 
      $status = $query->row()->account_status;} 
      else { 
      $status = ''; } 
      if($status == 'active') 
      { 
       $this->session->set_userdata($data); 
       redirect('site1/members_area'); 
      } 
      else //Account In active 
      { $this->inactive(); } 
     } 
     else // incorrect username or password 
     { 
      $this->invalid(); 
     } 
    } 
+0

嘗試echo $ status = $ query-> row() - > account_status;然後查看值 –

+0

我把代碼放在$ this-> inactive()的位置;當我登錄爲活動用戶時,它顯示爲非活動狀態,並且顯示無效用戶名。嘗試獲取非對象的屬性 – user2991258

+0

print_r($ query-> num_rows()); –

回答

0

嘗試運行這段代碼:

function validate_credentials() 
{  
    $this->load->model('membership_model'); 
    $query = $this->membership_model->validate(); 

    if($query) // if the user's credentials validated... 
    { 
     $data = array(
      'username' => $this->input->post('username'), 
      'is_logged_in' => true 
     ); 
     if($query->num_rows()>0) 
     { 
      $status = $query->row()->account_status; 
      if($status == 'active') 
      { 
       $this->session->set_userdata($data); 
       redirect('site1/members_area'); 
      } 
      else //Account In active 
      { $this->inactive(); } 
     } 
     else 
     { 
      $this->invalid(); 
     } 

    } 
    else // incorrect username or password 
    { 
     $this->invalid(); 
    } 
} 

,如果你仍然有問題給我查詢的話,我可以幫你。