[英]Submit a form and display the result in alertbox using opencart
我需要在opencart中显示警告框。 这意味着,一旦我提交了数据,我需要在警报框中显示结果
这是我的view.tpl文件
<script type="text/javascript">
$('#submit').submit(function() {
$.ajax({ // create an AJAX call...
data: $(this).serialize(),
type: $(this).attr('method'),
url: $(this).attr('action'),
success: function(response) {
$('#submit').html(response);
}
});
return false;
});
</script>
<form action="<?php echo $action; ?>" method="post">
<input type="text" name="your_name"><br>
<input type="text" name="email"><br>
<input type="submit" value="submit" name="submit">
</form>
这是我的控制器文件
public function index()
{
$this->load->model('test/test');
$this->data['action'] = $this->url->link('test/test');
$data['footer'] = $this->load->controller('common/footer');
$data['header'] = $this->load->controller('common/header');
$this->response->setOutput($this->load->view('test/test', $data));
if(isset($_POST['submit']) )
{
$your_name = $_POST["your_name"];
echo $your_name;
$email =$_POST["email"];
echo $email;
}
您需要在控制器中以JSON形式返回。
e.g return json_encode($var);
这样的事情。
public function index() {
$this->load->model('test/test');
$this->data['action'] = $this->url->link('test/test');
$data['footer'] = $this->load->controller('common/footer');
$data['header'] = $this->load->controller('common/header');
$this->response->setOutput($this->load->view('test/test', $data));
$result = [];
if( isset($_POST['submit']) ){
$your_name = $_POST["your_name"];
$email =$_POST["email"];
$result['yourname'] = $your_name;
$result['email'] = $email;
}
return json_encode($result);
}
然后, console.log(response);
在你的js中。
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