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