I have set up my Codeigniter application so that I can upload files via Ajax. I followed this tutorial http://net.tutsplus.com/tutorials/javascript-ajax/how-to-upload-files-with-codeigniter-and-ajax/
My original form checked to see if an ajax request had been called, if not then I had the fallback CI form validation / error messages showing instead.
I checked this using - $this->input->is_ajax_request()
My code looked like this:
if($this->input->is_ajax_request()){
// process ajax form data
} else {
if($this->form_validation->run() == FALSE) {
$data['success'] = 0;
$data['errors'] = validation_errors();
} else {
$data['success'] = 1;
}
$this->load->view('form', $data);
}
After doing some investigation I discovered that I couldn't apply the same technique because it isn't actually an ajax request, therefore I am not sure how I can use this approach. If anyone can point me in the right direction that would be great. I don't like it being totally dependent on ajax, I like having a fallback option. I noticed in the comments that someone has set up a CSFR cookie in their ajaxfileupload.js but to be honest I'm not too hot with js so I wouldn't know where to begin. Thanks in advance.
In your AJAX request along with everything else you could post key/value:
ajax : 1
Then in your controller:
if( $this->input->post('ajax') == 1 ) {
// process ajax form data
}
else
{
// form validation
}
Hope this helps.
Related
I am using Yii-user extension in the main layout i have a sign up link which is common to all the Cmenu
view/main layout
echo CHtml::link('Signup','#',array('id'=>'regi'));
$("#regi").click(function(){
$.ajax({
type:'GET',
url:'<?php echo Yii::app()->request->baseUrl;?>/index.php/user/registration',
success:function(res){
$("#dispdata").show();
$("#dispdata").html(res);
}
});
});
<div id="dispdata"><div>
**yii user extension **renders this perfectly and even submit its correctly if form values a re valid.
but if the values are incorrect and blank it redirect to url .../user/registration
which is not what my need .I need guidance what do i do such that if the values are incorrect or blank it should not redirect and display the errors in model window.
I did tried but hardly could get the satisfied results
if i place the following the model window itself doesnt appear what do i do
module registrationController i placed
....//some code here (**in yiiuser register controller**)
if ($model->save()) {
echo CJSON::encode(array(
'status'=>'success',
));
}
....//some code here...
Yii::app()->clientScript->scriptMap['jquery.js'] = false;
$this->renderPartial('registration',array('model'=>$model,),false,true);
in module view registration
<?php echo CHtml::ajaxSubmitButton(Yii::t('registration'),CHtml::normalizeUrl(array('user/registration','render'=>false)),array('dataType'=>'json',
'success'=>'function(data) {
if(data != null && data.status == "success") {
$("#registration-form").append(data.data);
}
}')); ?>
can anyone please guide me am working past 10 ten days tried every hook or crook method but could not obtain the results......how can the model window with client side validation be done appear..... Please guide me or let me know something better can be done
rules in registration model
if (!(isset($_POST['ajax']) && $_POST['ajax']==='registration-form')) {
array_push($rules,array('verifyCode', 'captcha', 'allowEmpty'=>!UserModule::doCaptcha('registration')));
as well was not with attributes for reqired field
have changed to
array_push($rules,array('verifyCode', 'captcha','message' => UserModule::t("captcha cannot be blank.")));
and added the verifycode to required field
yet not working,
The simple way is using render method in your Ajax action and creating empty layout for this action. If you do so, validation scripts will be included in the server response. Also you need to exclude jquery.js and other script with Yii::app()->clientScript->scriptMap and include them in main layout always.
I have a problem repopulating my form after validation fails. Problem is my url contains an additional uri which I access by clicking a link. This is what it looks like:
http://www.example.com/admin/trivia/add/5
At first trouble was that the segment 4 of the uri completely disappeared, so even though the validation errors showed and the form was repopulated, I lost my added uri.
Then I found in another question that the solution was to set form open like this:
echo form_open(current_url());
Problem is now it isn't showing any validation errors and the form is not repopulated. Is there a way to achieve this?
This is what my controller looks like:
function add()
{
$data = array('id' => $this->uri->segment(4));
if($_POST)
{
$this->_processForm();
}
$this->load->view('admin/trivia_form', $data);
}
And inside _processForm() I got all the validation rules, error message and redirecting in case success.
[edit] Here is my _processForm() :
function _processForm()
{
$this->load->library('form_validation');
//validation rules go here
//validation error messages
$this->form_validation->set_rules($rules);
$this->form_validation->set_error_delimiters('<div style="color:red">', '</div>');
if ($this->form_validation->run())
{
//get input from form and assign it to array
//save data in DB with model
if($this->madmin->save_trivia($fields))
{
//if save is correct, then redirect
}
else
{
//if not show errors, no redirecting.
}
}//end if validation
}
To keep the same url, you can do all things in a same controller function.
In your controller
function add($id)
{
if($this->input->server('REQUEST_METHOD') === 'POST')// form submitted
{
// do form action code
// redirect if success
}
// do your actual stuff to load. you may get validation error in view file as usual if validation failed
}
to repopulate the form fields you are going to need to reset the field values when submitting it as exampled here and to meke it open the same page you can use redirect() function as bellow:
redirect('trivia/add/5','refresh');
i don't know what you are trying to do, but try this to repopulate the form with the values user entered
<?php
echo form_input('myfield',set_value('myfield'),'placeholder="xyz"');
?>
I have been using the trick below to return success response with AJAX:
//In controller
echo 'success';
//In Javascript
if(response == 'success'){
//redirect
window.location.href = '/users/profile/';
}
It works fine on localhost. But in web server, I got the error below everytime I want to redirect the page after success:
Cannot modify header information - headers already sent by (output started at ...
So yeah, I know it is caused by the echo before redirecting.
So, Is there proper way to return the success response? No need to be a message, just true or false is enough.
[EDIT]
By using exit('success'), it works fine, but is this the best way?
Thanks.
It is better to redirect from javascript.
//In controller
echo 'success'; exit;
//In Javascript
if(response == 'success'){
window.location = 'your-controller-action';
}
EDIT
You may have space before/after php Opening/Closing tags in controller and models. Remove all the closing tags from all controllers and models and any whitespace before opening tags. Then check the result.
I am developing application using Ajax and jsp.
My index.jsp page has HTML code for Login and some Ajax code in javascript. Ajax will call my another CheckLogin.jsp page
CheckLogin.jsp page checks on server for valid username and password. It returns "success" if it's valid otherwise will return message stating "username or password is not valid."
Now, when username and passwrod is valid, instead of success, I want to redirect the page to "Home.jsp" what should I do?
I am new to jsp. I appreciate your help on this.
JSP code gets run once on the server before it goes to the client, so JSP/JSTL cannot do a redirect following the success or otherwise of an AJAX call (without a full page refresh - which obviates the use of AJAX). You should to do the redirect with Javascript:
if (success) {
var successUrl = "Home.jsp"; // might be a good idea to return this URL in the successful AJAX call
window.location.href = successUrl;
}
On successful AJAX call/validation, the browser window will reload with the new URL (effectively a redirect).
Since I don't see your code, you can integrate this somewhere inside your validation :
<%
pageContext.forward("logged.jsp");
%>
function Edit() {
var allVals = $('#NEWFORMCampaignID').val();
if (allVals > 0) {
window.location = '/SMS/PrepareSMS?id='+allVals;
}
else
alert("Invalid campaign to Edit")
}
In order to redirect using Button click and pass some parameters, you can call the Controller Path(#RequestMapping(value="/SMS/PrepareSMS")) and then handle it there..
In my CakePHP app for my login method I do some different things for when a user submits a form via AJAX calls using if ($this->request->is('ajax'))
However I also want to allow the login method to be shown in a modal for quick login which again is an ajax call. But how do I detect the difference between the AJAX GET to show the form and then AJAX POST to do the actual login?
See below I can detect native get and posts but for ajax how do I detect the difference in CakePHP??? As it seems I can only detect an ajax event and not the type :/
NATIVE:
GET = if ($this->request->is('get'))
POST = if ($this->request->is('post'))
AJAX:
GET = if ($this->request->is('ajax'))
POST = if ($this->request->is('ajax'))
Thanks
Solution:
if ($this->request->is('get'))
{
if ($this->request->is('ajax'))
{
echo json_encode('ajax get'); exit;
}
else {
echo 'Normal get'; exit;
}
}
if ($this->request->is('post'))
{
if ($this->request->is('ajax'))
{
echo json_encode('ajax post'); exit;
}
else {
echo 'Normal post'; exit;
}
}
Not sure if I understand the question, but if the problem is that the form data can come in either as a POST or GET, the solution is to check whether the POST data is there. If it is, use POST, otherwise take the data from GET. (Or other way around.)
If the function should do different things depending on whether the form has been sent as POST or GET, then simply make two different functions in the controller.