Question

您好我正在使用ajax创建一个使用CodeIgniter的登录表单，但它不能正常运行，因为我想要。它被重定向到我正在通过ajax执行的页面，但不执行do_login函数。如果我点击登录而不输入任何字段，它不会显示所需的字段，而我已在do_login函数中给出了验证码。请为我提供适当的解决方案，以解决如何使用这些函数重定向ajax的问题。这是我的控制器user.php

     class User extends CI_Controller {


public function index()
{
    $this->load->view('login');
}



public function do_login(){
    $this->form_validation->set_rules('username','Username','required');
    $this->form_validation->set_rules('password','Password','required|callback_verifyUser');
if($this->form_validation->run()== false){
    echo "sorry";                                     // where to put this echo to make it work through ajax
    $this->load->view('login');
}
else{
    echo "loggedIn";                                  // where to put this echo to make it work through ajax
    $data = array(
    'username' => $this->input->post('username'),
    'is_logged_in'=>1
    );
    $this->session->set_userdata($data);

    redirect('user/members');

}
}
public function members()
{
    if($this->session->userdata('is_logged_in')){

    $this->load->view('home');
    }
    else{
        redirect('user/restricted');
    }
}
public function restricted()
{
    $this->load->view('restricted');
}

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

    $this->load->model('LoginModel');
    if($this->LoginModel->login($username, $password)){
        return true;
    }
    else{
        $this->form_validation->set_message('verifyUser','Invalid Email or Password: Please enter it correctly');
        return false;
    }
}
public function logout(){
    $this->session->sess_destroy();
    redirect('user/index');
}


    }
    ?>

这是我的查看文件login.php

    <html>
    <head>
<link rel="stylesheet" type="text/css" href="http://maxcdn.bootstrapcdn.com/bootstrap/3.3.6/css/bootstrap.min.css"/>
<script type="text/javascript" src="http://ajax.googleapis.com/ajax/libs/jquery/2.1.3/jquery.min.js"></script>
<script type="text/javascript" src="http://127.0.0.1/simple_login_comp/js/jquery-1.9.1.js"></script>
<script type="text/javascript">
$(document).ready(function(){
    $("#frm_login").submit(function(event){
        event.preventDefault();

$.ajax({
url: "http://127.0.0.1/simple_login_comp/index.php/user/do_login",  
type: "POST",
data: 
{
    username: $('#username').val(),
    password: $('#password').val()},
success: function(data) 
{
    if (data== 'loggedIn')
    {   
        alert("you are logged IN");
        //window.location.replace("http://127.0.0.1/simple_login_redirect/index.php/user/home");
        //window.location.href="http://127.0.0.1/simple_login_comp/index.php/user/members";
    }
    else if(data== 'sorry'){

        alert("sorry");
    }
    //else{
    //  alert(data);
    //}
}
    });
});
});
</script>
</head>
<body>
<div class="container">
        <div class="row">
            <div class="span12">
                <div class="well">
                    <center><h1>Be a member of Mrwebsiter</h1></center>
                </div>
            </div>
        </div>
<h1>Login</h1>
<p>Please fill your details to login</p>

<form action="" id="frm_login" method="post">
Username :</br>
<input type="text" name="username" id="username"/></br>
Password :</br>
<input type="password" name="password" id="password"/></br>

<input type="submit" value="Login" name ="submit"/>
</form>
<div class="row">
    <div class="span12">
        <div class="well">
            <center><h3>copyright</h3></center>
        </div>
    </div>
</div>
</div>
</body>
</html>

这是我的模型LoginModel.php

     <?php


    class LoginModel extends CI_Model {


public function login($username, $password)
{
    $this->db->select('username', 'password');
    $this->db->from('users');
    $this->db->where('username', $username);
    $this->db->where('password', $password);

    $query= $this->db->get();

    if($query->num_rows() == 1)
    {
        return true;
    }
    else
    {
        return false;
    }
}



}
?>

我的登录表单不起作用，任何人都可以建议我如何让它运行，我在哪里做错了。当我通过ajax重定向它时，它不会通过do_login函数验证输入，而是直接重定向到页面。

Answer 1

这是您的问题

 if($this->form_validation->run()== false){
        echo "sorry";                                     // where to put this echo to make it work through ajax
        $this->load->view('login');
    }

将此更改为

if($this->form_validation->run()== false){
     echo "sorry";exit
    }

Answer 2

我建议您在控制器上使用 echo json_encode（$ data）; ，其中 $ data 是包含您的require字段或任何提示的关联数组。

在你的ajax句柄上使用 -

进行回调

success: function(data) {
   var response = eval('(' + data + ')');
   // DOM codes here
} // rest of the codes here

如果你不熟悉这个并且你想在登录系统上推送这个ajax的东西，我建议你查看javascript dom，callbacks和其他相关内容。

Answer 3

在您的控制器上

public function do_login(){
    $this->form_validation->set_rules('username','Username','required');
    $this->form_validation->set_rules('password','Password','required|callback_verifyUser');
    if($this->form_validation->run()== false){
        echo "sorry";                                     // where to put this echo to make it work through ajax

    }
    else{

            $data = array(
             'username' => $this->input->post('username'),
             'is_logged_in'=>1
             );
            $this->session->set_userdata($data);
            echo "loggedIn";         // where to put this echo to make it work through ajax

     }
 }

在您的观看

<script type="text/javascript">
$(document).ready(function(){
$("#frm_login").submit(function(event){
    event.preventDefault();

$.ajax({
url: "http://127.0.0.1/simple_login_comp/index.php/user/do_login",  
type: "POST",
data: 
{
    username: $('#username').val(),
    password: $('#password').val()},
    success: function(data) 
    {
        if (data== 'loggedIn')
        {   
            alert("you are logged IN");
            window.location ="http://127.0.0.1/simple_login_redirect/index.php/user/members");

        }
        else if(data== 'sorry'){

            alert("sorry");
        }

    }
    });
});
});
</script>

CodeIgniter使用Ajax登录表单

3 个答案: