I have built a model from GET request and display the content that I need into a form, mainly dropdown options. User then completes the form and 'POST' back to the api. The API that I am using isn't formatted in a way that I can use for ember-data so I have opted to render my model with Ember.Object
var Prequalification = Ember.Object.extend();
Prequalification.reopenClass({
template: function(){
return Ember.$.ajax({
url: "/prequalification",
dataType: 'json'
}).then(function(response){
var template = response.collection.template.data;
return template;
});
}
});
export default Prequalification;
My controller decorates the view:
var IndexController = Ember.ArrayController.extend({
businessType: function(){
var content = this.get('content');
console.log(this);
return content.get(10);
}.property('content'),
loanType: function(){
var content = this.get('content');
return content.get(5);
}.property('content')
});
export default IndexController;
So on form submit, what are my options for Posting back to the API?
Thanks!
Use an action and associate it with a button.
actions: {
save: function(){
alert('ajax save here');
}
}
http://emberjs.jsbin.com/jositowa/1/edit
Related
I have trying to load a page using simple get request
<script>
$(document).ready(function(){
$("li").click(function(e){
e.preventDefault();
var href = $("a",this).attr('href');
$.ajax({
async: true,
type: "GET",
url: href,
success: function (response) {
$('#main-content').html(response);
}
})
});
});
</script>
In response I am getting content with full layout. But Here I am trying to get only content without layout. In controller I have written code like below
public function index()
{
if ($request->ajax())
{
$this->layout = null; //but same result
}
$tags = Tag::orderBy('id', 'desc')->paginate(10);
return view('admin.tags.index')->with('tags',$tags);
}
But I am getting same result with layout, how can I make layout false ? Or can change layout in controller ?
you can use the renderSections
return view('admin.tags.index')->renderSections()['content'];
https://laravel-tricks.com/tricks/render-view-without-layout
Well, you are still returning a view. Why not do:
return response()->json(["data"=>"some data"]);
after your Ajax check? This will return a JSON response to your frontend.
whenever I send a post request to 'tasks/add' i want the user to return to a new page, but all I get is plain html text in a popup.
Route.php code
Route::post('tasks/add', function() {
return view('secrets');
});
this is my ajax request :
$("#frm").submit(function(e){
e.preventDefault();
var customer = $("input[name=customer]").val();
var details = $("input[name=details]").val();
var dataString = 'customer='+customer+'&details='+details;
$.ajax({
url: "tasks/add",
type:"POST",
beforeSend: function (xhr) {
var token = $('meta[name="csrf_token"]').attr('content');
if (token) {
return xhr.setRequestHeader('X-CSRF-TOKEN', token);
}
},
data : dataString,
success:function(data){
console.log(dataString);
alert(data);
},error:function(){
alert("error!!!!");
}
}); //end of ajax
});
});
Anybody has had this issue before?
You are using Ajax to call your Route method. So when Route::post(...) returns the view 'secrets', it returns it to the Ajax method and becomes held by the data variable. Saying return in your Routes file doesn't magically mean redirect to a certain view, it is just like any other function that returns a value.
You currently have alert(data) which just says make an alert with whatever is held by data which in this case is the html text of your view.
Instead, take out the alert() and put
window.location.replace('<path/to/secrets>')
to redirect to the page you want upon success.
Assuming your Routes file has something like :
Route::get('/secrets', function() {
return view('secrets');
});
You could say
window.location.replace('/secrets')
I have following js code:
window.addEvent('domready', function() {
var trigger = $('sendme');
trigger.addEvent( 'click', function(event){
event.preventDefault()
var sendform = new Form.Request($('newform'), {
onSend: function(){
console.log('sending');
},
onComplete: function(){
console.log('sent');
}
});
sendform.send();
});
});
and form with data:
<form action="index.php?option=com_mycomp&layout=edit&id=1" method="post" name="newform" id="newform" class="form-validate">...
the form submits just fine and I can see changes but I get no logs,
thus cant execute actions that I need
form action is not supposed to give me any response back , it is simple post but shouldn't this work? Do I need to send the form to another file that will give me responses like json and submit my form like that ?
what am I doing wrong ?
Any help is appreciated. Thnx!
small update since post ,
I change the form to send data and receive response via json file but still no response messages. everything is being updated so submit works 100%.
right way is new Form.Request($('newform'),console.log(),{
window.addEvent('domready', function() {
var trigger = $('sendme');
trigger.addEvent( 'click', function(event){
event.preventDefault()
var sendform = new Form.Request($('newform'),console.log(), {
onSend: function(){
console.log('sending');
},
onComplete: function(){
console.log('sent');
}
});
sendform.send();
});
});
I have made a custom module in magento. I am using ajax in it(prototype.js).i can find the post variable on ajax page. But I am unable to find the file array on ajax page.
I am using following code for this.Please let me know where i am wrong?
//Ajax code on phtml page
new Ajax.Request(
reloadurl,
{
method: 'post',
parameters: $('use-credit-Form').serialize(),
onComplete: function(data)
{
alert(data.responseText);
}
});
//Php code on ajaxpage
public function ajaxAction()
{
$fileData = $_FILES;
echo '<pre>';
print_r($fileData);die;
}
It always print blank. but when I added this line
"VarienForm.prototype.submit.bind(usecreditForm)();"
I can get the value of file array. but draw back now page starts refreshing.
Please give me some suggestion.
Try this:
Event.observe('use-credit-Form', 'submit', function (event) {
$('use-credit-Form').request({
onFailure: function () {
alert('fail.');
},
onSuccess: function (data) {
alert(data.responseText);
}
});
Event.stop(event); // stop the form from submitting
});
Credit: submit a form via Ajax using prototype and update a result div
I'm trying to use Ajax in CakePHP, and not really getting anywhere!
I have a page with a series of buttons - clicking one of these should show specific content on the current page. It's important that the page doesn't reload, because it'll be displaying a movie, and I don't want the movie to reset.
There are a few different buttons with different content for each; this content is potentially quite large, so I don't want to have to load it in until it's needed.
Normally I would do this via jQuery, but I can't get it to work in CakePHP.
So far I have:
In the view, the button control is like this:
$this->Html->link($this->Html->image('FilmViewer/notes_link.png', array('alt' => __('LinkNotes', true), 'onclick' => 'showNotebook("filmNotebook");')), array(), array('escape' => false));
Below this there is a div called "filmNotebook" which is where I'd like the new content to show.
In my functions.js file (in webroot/scripts) I have this function:
function showNotebook(divId) {
// Find div to load content to
var bookDiv = document.getElementById(divId);
if(!bookDiv) return false;
$.ajax({
url: "ajax/getgrammar",
type: "POST",
success: function(data) {
bookDiv.innerHTML = data;
}
});
return true;
}
In order to generate plain content which would get shown in the div, I set the following in routes.php:
Router::connect('/ajax/getgrammar', array('controller' => 'films', 'action' => 'getgrammar'));
In films_controller.php, the function getgrammar is:
function getgrammar() {
$this->layout = 'ajax';
$this->render('ajax');
}
The layout file just has:
and currently the view ajax.ctp is just:
<div id="grammarBook">
Here's the result
</div>
The problem is that when I click the button, I get the default layout (so it's like a page appears within my page), with the films index page in it. It's as if it's not finding the correct action in films_controller.php
I've done everything suggested in the CakePHP manual (http://book.cakephp.org/view/1594/Using-a-specific-Javascript-engine).
What am I doing wrong? I'm open to suggestions of better ways to do this, but I'd also like to know how the Ajax should work, for future reference.
everything you show seems fine. Double check that the ajax layout is there, because if it's not there, the default layout will be used. Use firebug and log function in cake to check if things go as you plan.
A few more suggestions: why do you need to POST to 'ajax/getgrammar' then redirect it to 'films/getgrammar'? And then render ajax.ctp view? It seems redundant to me. You can make the ajax call to 'films/getgrammar', and you don't need the Router rule. You can change ajax.ctp to getgrammar.ctp, and you won't need $this->render('ajax');
this is ajax call
$(function() {
$( "#element", this ).keyup(function( event ) {
if( $(this).val().length >= 4 ) {
$.ajax({
url: '/clients/index/' + escape( $(this).val() ),
cache: false,
type: 'GET',
dataType: 'HTML',
success: function (clients) {
$('#clients').html(clients);
}
});
}
});
});
This the action called by ajax
public function index($searchterm=NULL) {
if ( $this->RequestHandler->isAjax() ) {
$clients=$this->Client->find('list', array(
'conditions'=>array('LOWER(Client.lname) LIKE \''.$searchterm.'%\''),
'limit'=>500
));
$this->set('clients', $clients);
}
}
This is a function I use to submit forms in cakephp 3.x it uses sweet alerts but that can be changed to a normal alert. It's very variable simply put an action in your controller to catch the form submission. Also the location reload will reload the data to give the user immediate feedback. That can be taken out.
$('#myForm').submit(function(e) {
// Catch form submit
e.preventDefault();
$form = $(this);
// console.log($form);
// Get form data
$form_data = $form.serialize();
$form_action = $form.attr('action') + '.json';
// Do ajax post to cake add function instead
$.ajax({
type : "PUT",
url : $form_action,
data : $form_data,
success: function(data) {
swal({
title: "Updated!",
text: "Your entity was updated successfully",
type: "success"
},
function(){
location.reload(true);
});
}
});
});