Question

我基本上是将表单值提交给我的控制器，而我正在使用CodeIgniter Framework。但是，当我将值发送到我的控制器的函数时，页面将更改为控制器并离开index.php（当前页面）

的index.php：

<form action="<?php echo base_url();?>index.php/LoginController/loginuser" method="post">
            <input id="login_emailbox" name="login_emailbox" type="text" class="form-control welcome-login-email" placeholder="Email" required="">
            <input id="login_passbox" name="login_passbox" type="password" class="form-control welcome-login-password" placeholder="Password" required="">
            <button id="loginbtn" type="submit" class="btn btn-info" style="margin-left: 30px">Login</button>
</form>

LoginController.php：

class LoginController extends CI_Controller {

    public function __construct()
    {
        parent::__construct();
    }

    function index()
    {
        $this->load->helper('html');
        $this->load->helper('url');
        $this->load->view('header');
        $this->load->view('footer');

        $this->load->view('index.php');
    }


    public function loginuser(){
        echo $_POST['login_emailbox'];
        echo $_POST['login_passbox'];
    }

}

我在运行时采取的步骤：

1）我浏览到index.php via =＆gt; http://localhost/codeig/index.php/LoginController/index

2）填写表格并点击提交。值将提交给函数： loginuser

3）页面被重定向到'loginuser'函数

如何避免这种情况并基本上将值发送到控制器中的loginuser函数而不刷新当前页面？

Answer 1

<form action="" method="post" id="myForm">
            <input id="login_emailbox" name="login_emailbox" type="text" class="form-control welcome-login-email" placeholder="Email" required="">
            <input id="login_passbox" name="login_passbox" type="password" class="form-control welcome-login-password" placeholder="Password" required="">
            <button id="loginbtn" type="submit" class="btn btn-info" style="margin-left: 30px">Login</button>
</form>

你的Jquery

 $('#loginbtn').on('click',function(e){
e.preventDefault();
    var data = $('#myForm').serialize();
    var base_url='<?php echo base_url(); ?>'
    $.ajax({
    url:base_url+'index.php/LoginController/loginuser',
    type:'POST',
    data:data,
    success:function(data){

    alert(data); // here what you want to do with response
    }
    }); 
return false;


   });

Answer 2

    <form>
                    <input id="login_emailbox" name="login_emailbox" type="text" class="form-control welcome-login-email" placeholder="Email" required="">
                    <input id="login_passbox" name="login_passbox" type="password" class="form-control welcome-login-password" placeholder="Password" required="">
                    <button id="loginbtn" type="submit" class="btn btn-info" style="margin-left: 30px">Login</button>
    </form>

<script>
$(document).ready(function(){
              $('#loginbtn').click(function(){
                    var login_emailbox = document.getElementById('login_emailbox').value;

                    $.ajax({
                        url:'<?=base_url()?>index.php/Controller/function',
                        method: 'post',
                        data: {login_emailbox: login_emailbox},
                        dataType: 'json',
                        success: function(response){
alert('data updated'); 
                        }
                    });
                }); 
});
</script>

CodeIgniter：提交表单后避免刷新？

2 个答案: