Display form via Ajax - ajax

Edit
Many questions on SO are about form submission via Ajax (this one is very popular) but not about how to display a form through an ajax call.
My question is about the part in bold: the user asks for a form -> ajaxCall -> displayForm -> user submit the form -> send the form to the controller -> validates form -> send response to the user (errors or not).
The question
The ajax call being a POST, $request->isMethod('POST') always returns true when doing an ajax call. Therefore if a new form is being requested through an ajax call, it will always be validated. I have enclosed the code below, the form is displayed ok, but the following error message is displayed in the form:
The CSRF token is invalid.
The problem is that the form is being validated during the ajax call requestion the form in the controller.
On a side note, submitting the form works well, and validation works well.
Does anyone know how to display a form via Ajax without having the blank new form being validated?
My code so far:
/**
* Display the message as well as the form to reply to that message
*
* #param Request $request
* #return Response
*/
public function viewMessageAction(Request $request)
{
//Forbid every request but jQuery's XHR
if (!$request->isXmlHttpRequest()){
return new Response('', 200, array('Content-Type' => 'application/json'));
}
//Retrieve the message from the request
$messageId = $request->request->get('messageId');
$message = $this->getMessageManager->getMessage($messageId);
//Creates the form to reply to the message
$form = $this->container->get('reply_form.factory')->create($message);
if ($request->isMethod('POST')) {
$form->bind($request);
if ($form->isValid()) {
//...
return $this->redirect($this->generateUrl('task_success'));
}
}
$answer =$this->container->get('templating')->renderResponse('AcmeMessageBundle:Message:message.html.twig', array(
'form' => $form->createView(),
'message' => $message
));
$response = new Response();
$response->headers->set('Content-type', 'application/json; charset=utf-8');
$response->setContent(json_encode($answer));
return $response;
}
EDIT: jQuery part:
<script type="text/javascript">
$(function(){
var myPath = ;//...
function getMessage(messageId){
$.ajax({
type: "POST",
url: myPath,
data: "messageId="+messageId,
success: function(returnData){
$('#message').html(returnData);
}
});
}
});
</script>

An ajax call doesnt always need to be POST. It could be GET, PUT or DELETE . So when you are making ajax call to get the form make it a GET request.
If you are using jquery here is how you do that
$.ajax({
url: "http://localhost/proj/url",
type:'GET'
})
If you are using Using XHR, you can specify the type in the open function
xhr.open( 'GET', URL)
PS: It will be a good idea to use JMS Serializer Service for serialization instead of json_encode

You might also be interessted in the FOSJsRoutingBundle, since it add a lot of comfort working with ajax and Symfony2 routes in general.

Related

Insert Data Using Ajax

I am using Laravel 5.3. I want to insert the data using blade template.But my when i press submit button it gets refreshed every time. what to do? and please anyone tell me how to use ajax url,type,data
If you try to submit via Javascript make sure prevent form default action with e.preventDefault(). This code prevent the form submitted in a regular way. Just add this code to wrap your AJAX call:
$('#form-id').submit(function(e){
e.preventDefault();
$.ajax({...});
});
I just assume you are using jquery if you are talking about ajax. It's really simple. Your laravel routes listen to "post", "get", "patch", "delete" methods.
Everything of these can be created with a ajax request - example:
$.ajax({
method: "POST",
url: "/posts",
data: { title: "Hello World", text: "..." }
})
.done(function( post ) {
// assuming you return the post
alert(post.title + " created");
});
Now that you use ajax you will not want to return a view to the ajax call. You have different options here (create a new route, helper functions etc.) I will give the most easy example
Controller function:
public function store(Request $request) {
$post = App\Post::create($request->all());
if($request->ajax()) {
return $post;
} else {
return redirect('/posts');
}
}
now you controller will return data on ajax calls and will redirect you on default calls without ajax.
Finally you have a last thing to keep in mind. If you have web middleware applied ( done by default ) you need to handle the csrf token. The most easy way to handle this is by adding a meta tag to your html head
and then (before doing all your calls etc.) add this to configure your ajax
$.ajaxSetup({
headers: {
'X-CSRF-TOKEN': $('meta[name="_token"]').attr('content')
}
});
this will add the valid csrf token which is in your head to every ajax call and will ensure you not run into token missmatch exceptions.
Things to keep in mind:
- if you stay very long on one page tokens might expire ( laravel-caffeine will help here )
- you need to handle validation for ajax calls

Before redirect set validtion error to view page

i am calling the form from ajax and after submitting the form, validation error message not showing, ajax call on load.
my ajax code
$.ajax({
type:"POST",
url:"/application/workpermit",
data:dataString,
success:function(data){
$('#tabs-12 .workpermit').html(data);
//alert(data); return false;
}
});
Success return the workpermit view with form
The controller part, after submitting
if ($validator->fails()) {
return redirect("application/".$employer_maid_id."/edit?tab=tab11")
->withErrors($validator)
->withInput();
}
Please any one suggest me, how to show the validation message
if you are not refreshing the page you need to add(append) the message with js/jQuery
if you are redirecting from the page you can you Laravel's built in feature "Flash". Basically it's a one time Session laravel docs

Refresh form in Django without reloading page

Hi I'm new in Ajax and django and I want to refresh my form. I try some code but it didn't work. I'm sure what I want to do is very basic.
Here my html:
<div class="row" style="padding-top:20px;">
<div class="col-md-12" id="testAjax">
{% load crispy_forms_tags %}
{% crispy form %}
</div>
</div>
I want to refresh my form in the div testAjax.
Here my view:
def createPin(request):
error = False
if request.method == "POST":
form = CreatePinForm(request.POST)
if form.is_valid():
pin = form.save(commit=False)
pin.customer = request.user.customer
pin.save()
msg = "pin saved"
return redirect('/pin/CreatePin', {'form': form, 'msg': msg})
else:
error = True
else:
form = CreatePinForm()
return render(request, 'createPin.html', {'form': form, 'error': error,})
My Ajax:
function refresh()
{
$form=$('#createPin');
var datastring = $form.serialize();
$.ajax({
type: "POST",
url: '/pin/CreatePin/',
dataType: 'html',
data: datastring,
success: function(result)
{
/* The div contains now the updated form */
$('#testAjax').html(result);
}
});
}
Thanks alot for your help.
When I need to do some operations and I don't want to reload the page I use a JQuery call to Ajax, I make the pertinent operations in AJAX and then receive the AJAX response in the JQuery function without leaving or reloading the page. I'll make an easy example here for you to understand the basics of this:
JQuery function, placed in the template you need
function form_post(){
//You have to get in this code the values you need to work with, for example:
var datastring = $form.serialize();
$.ajax({ //Call ajax function sending the option loaded
url: "/ajax_url/", //This is the url of the ajax view where you make the search
type: 'POST',
data: datastring,
success: function(response) {
result = JSON.parse(response); // Get the results sended from ajax to here
if (result.error) { // If the function fails
// Error
alert(result.error_text);
} else { // Success
//Here do whatever you need with the result;
}
}
}
});
}
You have to realize that I cannot finish the code without knowing what kind of results you're getting or how do you want to display them, so you need to retouch this code on your needs.
AJAX function called by JQuery
Remember you need to add an url for this Ajax function in your urls.py something like:
url(r'^/ajax_url/?$', 'your_project.ajax.ajax_view', name='ajax_view'),
Then your AJAX function, it's like a normal Django View, but add this function into ajax.py from django.core.context_processors import csrf from django.views.decorators.csrf import csrf_exempt from django.utils import simplejson
#csrf_exempt
def ajax_view(request):
response = []
#Here you have to enter code here
#to receive the data (datastring) you send here by POST
#Do the operations you need with the form information
#Add the data you need to send back to a list/dictionary like response
#And return it to the JQuery function `enter code here`(simplejson.dumps is to convert to JSON)
return HttpResponse(simplejson.dumps(response))
So, without leaving the page you receive via javascript a list of items that you sended from ajax view.
So you can update the form, or any tag you need using JQuery
I know that this can be so confusing at the beginning but once you are used to AJAX this kind of operations without leaving or reloading the page are easy to do.
The basics for understanding is something like:
JQuery function called on click or any event you need
JQuery get some values on the template and send them to AJAX via
POST
Receive that information in AJAX via POST
Do whatever you need in AJAX like a normal DJango view
Convert your result to JSON and send back to the JQuery function
JQuery function receive the results from AJAX and you can do
whatever you need

How to avoid redirect after form submission if you have a URL in your form's action?

I have a form that looks like this:
<form name="formi" method="post" action="http://domain.name/folder/UserSignUp?f=111222&postMethod=HTML&m=0&j=MAS2" style="display:none">
...
<button type="submit" class="moreinfo-send moreinfo-button" tabindex="1006">Subscribe</button>
In the script file I have this code segment where I submit the datas, while in a modal box I say thank you for the subscribers after they passed the validation.
function () {
$.ajax({
url: 'data/moreinfo.php',
data: $('#moreinfo-container form').serialize() + '&action=send',
type: 'post',
cache: false,
dataType: 'html',
success: function (data) {
$('#moreinfo-container .moreinfo-loading').fadeOut(200, function () {
$('form[name=formi]').submit();
$('#moreinfo-container .moreinfo-title').html('Thank you!');
msg.html(data).fadeIn(200);
});
},
Unfortunately, after I submit the datas, I'm navigated to the domain given in the form's action. I tried to insert return false; in the code (first into the form tag, then into the js code) but then the datas were not inserted into the database. What do I need to do if I just want to post the data and stay on my site and give my own feedback.
I edited Eric Martin's SimpleModal Contact Form, so if more code would be necessary to solve my problem, you can check the original here: http://www.ericmmartin.com/projects/simplemodal-demos/ (Contact Form)
Usually returning false is enough to prevent form submission, so double check your code. It should be something like this
$('form[name="formi"]').submit(function() {
$.ajax(...); // do your ajax call here
return false; // this prevent form submission
});
Update
Here is the full answer to your comment
I tried this, but it didn't work. I need to submit the data in the succes part, no?
Maybe, it depends from your logic and your exact needs. Normally to do what you asking for I use the jQuery Form Plugin which handle this kind of behavior pretty well.
From your comment I see that you're not submitting the form itself with the $.ajax call, but you retrieve some kind of data from that call, isn't it? Then you have two choices here:
With plain jQuery (no form plugin)
$('form[name="formi"]').submit(function() {
$.ajax(...); // your existing ajax call
// this will post the form using ajax
$.post($(this).attr('action'), { /* pass here form data */ }, function(data) {
// here you have server response from form submission in data
})
// this prevent form submission
return false;
});
With form plugin it's the same, but you don't have to handle form data retrieval (the commented part above) and return false, because the plugin handle this for you. The code would be
$(document).ready(function() {
// bind 'myForm' and provide a simple callback function
$(form[name="formi"]).ajaxForm(function() {
// this call back is executed when the form is submitted with success
$.ajax(...); // your existing ajax call
});
});
That's it. Keep in mind that with the above code your existing ajax call will be executed after the form submission. So if this is a problem for your needs, you should change the code above and use the alternative ajaxForm call which accepts an options object. So the above code could be rewritten as
$(document).ready(function() {
// bind 'myForm' and provide a simple callback function
$(form[name="formi"]).ajaxForm({
beforeSubmit: function() { $.ajax(...); /* your existing ajax call */},
success: function(data) { /* handle form success here if you need that */ }
});
});

get $.post to work with the validate plugin on multiple forms without seperate functions

On a fansite im doing http://yamikowebs.com/ee/
I have a few forms (2 atm). I used $.post to find out what form is being submited. submit the form and display that pages results where the form was originally with .html().
My next step was to use the validator which is working fine but im not sure how to put the 2 together.
submitHandler: function(form){} seems to be the setting for how its submitted. However, I can't get this to work with my $.post function or find out what form is being processed.
If I leave the defaults for validation plug-in if there no errors it will send you to the page. the ajax plug-in that it works with doesn't do what I want. Below is my $.post function
form validation:
//ajax post
$("form").submit(function(event)
{
event.preventDefault();//stop from submiting
//set needed variables
var $form = $(this)
var $div = $form.parent("div")
$url = $form.attr("action");
//submit via post and put results in div
$.post( $url, $form.serialize() , function(data)
{ $div.html(data) })
})
http://docs.jquery.com/Plugins/validation#source is the validation plugin
You're correct in thinking that submitHandler is the right callback to use. However, I ran into some interesting issues while using it with multiple forms (like you're trying to do). For example, in this code:
$("#form1, #form2").validate({
submitHandler: function(form) {
alert(form.action);
alert(form.id);
}
});
The submitHandler callback does not get supplied the correct parameter (it always gets #form1). I believe this is actually a bug in jQuery-validate (so I've filed it here).
Anyway, a decent workaround would be to wrap the validate call in .each():
$("form").each(function() {
$(this).validate({
submitHandler: function(form) {
/* 'form' has the correct value */
var values = $(form).serialize(),
$div = $(form).parent("div");
alert(form.action);
alert(form.id);
/* Perform AJAX call here */
}
});
});
Example: http://jsfiddle.net/andrewwhitaker/MmCXN/

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