开始之前，对不起我的英语。我正在使用后端上的 CI 开发一个网站。我想建立一个注册系统而不刷新页面。如果我尝试提交表单，我没有错误，一切都很好。但是当我尝试与ajax request一起使用时，因为表单验证返回false并且validation_errors为empty，所以我无法使用表单验证。如果我禁用表单验证，则ajax请求效果很好。这是我的控制器和ajax请求。请帮助我。
User.php（我的控制器）
public function register(){
$this->load->library('form_validation');
$this->form_validation->set_rules('email_kyt', 'Email', 'is_unique[users.email]');
$this->form_validation->set_rules('username_kyt', 'Kullanici', 'is_unique[users.username]');
if($this->form_validation->run() == FALSE) {
$data = json_encode(array('status'=> false,'info'=>validation_errors()));
}else {
if($this -> input -> is_ajax_request()){
$userData = array(
'email' => strip_tags($this->input->get('email_kyt')),
//bla bla,
);
if ($this->User_model->form_insert($userData) == true) { //this method works perfectly.
$data = json_encode(array('status' => true,'info' => 'Successfully Registered'));
} else {
$data = json_encode(array('status' => false,'info'=>'The Error Occurred During Registration'));
}
}else{
$data = json_encode(array('status'=> false,'info'=>'This is not Ajax request'));
}
}
echo $data;
}
}
这是我在 js
中的ajax请求
$(document).ready(function(){
$('#btn_register').on('click',function (e) {
$('form[name=register-form]').unbind("submit");
$('form[name=register-form]').submit(function (e) {
e.preventDefault();
$.ajax({
type: 'get',
url: url + 'User/register', //url is correct i tested without form validation
data: $('#register-form').serialize(),
dataType: "json",
success: function (data) {
if (data.status == true) {
alert(data.info);
window.location.reload();
json= [];
} else if (data.status == false) {
$('#span_validate').html(data.info);
json= [];
}
}
});
});
});
});
编辑：这是我的表格：
<!-- Register Form -->
<?php echo form_open(base_url('index.php/User/register'),array('id' => 'register-form','name' => 'register-form')); ?>
<?php echo form_hidden($this->security->get_csrf_token_name(), $this->security->get_csrf_hash()); ?>
<div class="md-form">
<input type="text" id="name_kyt" name="name_kyt" class="form-control" data-toggle="popover" data-trigger="focus" data-content="...">
<label for="name_kyt" >Name</label>
</div>
<div class="md-form">
<input type="text" id="surname_kyt" name="surname_kyt" class="form-control" data-toggle="popover" data-trigger="focus" data-content="...">
<label for="surname_kyt" >Surname</label>
</div>
<div class="md-form">
<input type="text" id="username_kyt" name="username_kyt" class="form-control" data-toggle="popover" data-trigger="focus" data-content="...">
<label for="username_kyt" > Username </label>
</div>
<div class="md-form">
<input type="email" id="email_kyt" name="email_kyt" class="form-control" data-toggle="popover" data-trigger="focus" data-content="...">
<label for="email_kyt" >Email</label>
</div>
<div class="md-form">
<input type="password" id="password_kyt" name="password_kyt" class="form-control" data-toggle="popover" data-trigger="focus" data-content="...">
<label for="password_kyt" >Password</label>
</div>
<div class="md-form">
<input type="password" id="password_confirm" name="password_onay" class="form-control">
<label for="password_confirm" >Password Confirm</label>
</div>
<div class="form-group text-center">
<div class="row">
<div class="col-sm-10 col-sm-offset-3 mr-auto ml-auto">
<input type="submit" name="btn_register" id="btn_register" tabindex="4" class="btn btn-register mr-auto ml-auto" value="Register">
<p><span id="span_validate" class="label label-default mr-auto ml-auto"></span></p>
</div>
</div>
</div>
<!-- End of Register Form -->
<?php echo form_close(); ?>
答案 0 :(得分：0)
Please refer below example to validate a form in CodeIgniter using ajax call.
1. ajax code.
$(document).ready(function(){
$('#btn_register').on('click',function (e) {
$('form[name=register-form]').unbind("submit");
$('form[name=register-form]').submit(function (e) {
e.preventDefault();
var formData = $("#register-form").serialize();
$.ajax({
type: 'get',
url: url + 'User/register',
data: formData,
success: function (data) {
if (data.status == true) {
alert(data.info);
window.location.reload();
json= [];
} else if (data.status == false) {
$('#span_validate').html(data.info);
json= [];
}
}
});
});
});
});
2. Controller code :
Load form_validation library and form helper
$this->load->library('form_validation');
$this->load->helper('form');
Now write your controller as ...
public function register(){
$this->load->library('form_validation');
$this->load->helper('form');
$this->form_validation->set_rules('email_kyt', 'Email', 'is_unique[users.email]');
$this->form_validation->set_rules('username_kyt', 'Kullanici', 'is_unique[users.username]');
if($this->form_validation->run() == FALSE) {
echo $data = json_encode(array('status'=> false,'info'=>validation_errors())); die;
}else {
if($this -> input -> is_ajax_request()){
$userData = array(
'email' => strip_tags($this->input->get('email_kyt')),
//bla bla,
);
if ($this->User_model->form_insert($userData) == true) { //this method works perfectly.
echo $data = json_encode(array('status' => true,'info' => 'Successfully Registered')); die;
} else {
echo $data = json_encode(array('status' => false,'info'=>'The Error Occurred During Registration')); die;
}
}else{
echo $data = json_encode(array('status'=> false,'info'=>'This is not Ajax request')); die;
}
}
}
}
答案 1 :(得分：0)
我已经解决了这个问题。表单验证仅适用于post方法。如果使用get方法，它将无法正常工作。