Question

我在codeigniter网站的firebug中收到此错误无法访问与您的字段名称密码对应的错误消息我已经搜索过，发现当代码中有错误时会弹出错误，但是没有像我这样的问题。你能帮忙吗？

登录控制器

function do_login()
{
 $username = $this->input->post('username');
 $password = $this->input->post('password');

 if(!empty($username) && isset($username))
 {
 $this->form_validation->set_rules(
      'username', 'Your Username',
      'trim|required|xss_clean|max_length[20]|callback_username_check');

 $this->form_validation->set_rules('password', 'Password','trim|required||xss_clean|callback_password_check['.$username.']|min_length[4]|max_length[20]');

if($this->form_validation->run() == FALSE)
{

 echo json_encode(array('st'=>0, 'msg' => validation_errors()));
}
 else if(($this->form_validation->run() == TRUE))
{ 
echo json_encode(array('st'=>1));
}
function username_check($username)
{
$this->form_validation->set_message('username_check', ' is not regesterd!');
$checkVar=$this->user_model->check_user($username);
if($checkVar == true)
{return TRUE;}
elseif ($checkVar == false)
{return false;}
}   
function password_check($username,$password)
{
$this->form_validation->set_message('password_check', ' is not correct!');
$checkVar =  $this->user_model->check_pass($username,$password);
if($checkVar == true)
{return TRUE;}
elseif ($checkVar == false)
{return false;
}}}}

user_model

public function check_user($username) 
{
$this->query = $this->db->select('COUNT(*)')
    ->from('users')
    ->where(array('username'=>$username))
    ->limit(1)->get();
$query = $this->query->num_rows();
if ($query> 0){
    return true;
}
else{
    return false;
}
}
 public function check_pass($username, $password) 
{
$this->query = $this->db->select('COUNT(*)')
    ->from('users')
    ->where(array('username'=>$username, $password))
    ->limit(1)->get();
$query = $this->query->num_rows();
if ($query> 0){
    return true;
}
else{
    return false;
}
}

AJAX

<script type="text/javascript">
$(document).ready(function() {
$('#frm').submit(function(){

$.post(
     $('#frm').attr('action'), 
     $('#frm').serialize(), 
     function( data ) 
     {
     if (data.st == 0)
     {
     $('#validation-error').html(data.msg);
     }
     else if(data.st == 1)
     {           var url = "home/"
     window.location.href = url;

     }
     }, 
     'json'
   );
 return false;   
  });


});

</script>

形式

<?php echo form_open('signin/do_login', array('id'=>'frm')); ?>

                                <input type="text" id='username' name="username" size="15" placeholder="اسم المستخدم">                              

                                                                    <input type="password" id="password" name="password" size="15" placeholder="كلمة المرور" >                              
                                                                        <div id="validation-error"></div>
                            <input type='submit' value="تسجيل الدخول">
                                        </form>

Answer 1

搜索和调试后我发现问题出现在这一行

 $this->form_validation->set_rules('password', 'Password','trim|required||xss_clean|callback_password_check['.$username.']|min_length[4]|max_length[20]');

应该是

 $this->form_validation->set_rules('password', 'Password','trim|required|xss_clean|callback_password_check['.$username.']|min_length[4]|max_length[20]');

问题是＆＃34; | ＆＃34;介于必需和xss之间谢谢

Answer 2

你在控制器

中失去了方括号

'trim|required||xss_clean|callback_password_check['.$username.'|min_length[4]|max_length[20]'

应该是

'trim|required||xss_clean|callback_password_check['.$username.']|min_length[4]|max_length[20]'

表单验证：无法访问与您的字段名称密码对应的错误消息

2 个答案: