Jquery ajax call in Cake PHP 3.0 - javascript

Hi,
I'm trying to make a ajax request to the view A from the controller B like this :
In the view A :
var tab = new Array();
function updateResult(){
$.ajax({
type:"POST",
url:"<?php echo Router::url(array('controller'=>'B','action'=>'viewresult'));?>",
dataType: 'text',
async:false,
success: function(tab){
alert('success');
},
error: function (tab) {
alert('error');
}
});
}
In the controller B:
public function viewresult()
{
echo 'SUCCESS';
}
The problem is that in the 'response' of ajax, I've 'SUCCESS' but also the entire view A, I don't understand why...
I want only 'SUCCESS'...
Thanks in advance !

Detect if its ajax as per following code in cakephp way :
if($this->request->is('Ajax')) //Ajax Detection
{
$this->autoRender = false; // Set Render False
$this->response->body('Success');
return $this->response;
}
Check here for more detectors - http://book.cakephp.org/3.0/en/controllers/request-response.html#Cake\Network\Request::addDetector
You can also use $this->Url->build instead of including Router for creating links in view.
echo $this->Url->build(['action'=>'index']);

The easiest way to achieve it is adding die() at the end of your function so it prevents to load whole layout:
public function viewresult()
{
echo 'SUCCESS';
die;
}
OR
public function viewresult()
{
die('SUCCESS');
}
But more conventional way is using JSONView. Your action should look as follows:
public function viewresult()
{
$this->set('text', 'SUCCESS');
$this->set('_serialize', ['text']);
}
You also have to load RequestHandler component in initialize() method in your controller:
public function initialize()
{
parent::initialize();
$this->loadComponent('RequestHandler');
}
You need to set allowed extensions for all routes connected later in routes.php:
Router::extensions('json', 'xml');
Now you can access your action adding extension .json at the end of it's URL, so you need to modify ajax call url:
url:"<?php echo Router::url(array('controller'=>'Main','action'=>'viewresult', '_ext' => 'json'));?>"
That's all, but keep in mind that this solution force you to handle JSON array in response. In this example output will be looks as follows:
{"text": "SUCCESS"}

Related

Sending data of ajax by another ajax to controller in laravel

I want to know is that possible to send data from one ajax by another ajax or not?
Sounds confusing I know, but here is explanation:
I have payment method where it gets data and handling them by Ajax (unfortunately the creators of this API limited their code a lot) so even if i try to add input request in controller code of that Ajax nothing will work, that's why I need to make another Ajax to handle that input request.
Lets explain more by codes:
controller
public function orderspayonline(Request $request, $id){
error_log('masuk ke snap token dri ajax');
$midtrans = new Midtrans;
//products data + user info etc.
//here magic happens
try
{
$snap_token = $midtrans->getSnapToken($transaction_data);
echo $snap_token;
}
catch (Exception $e)
{
return $e->getMessage;
}
}
If I add anything (I mean anything) in that try{ part it will stop functioning and return error! `even I tried to redirect back my users except echoing token code that gave error as well. So it seems I really don't have any option here but to create new function and Ajax.
JavaScript
<script type="text/javascript">
$('.pay-button').click(function (event) {
$.ajaxSetup({
headers: { 'X-CSRF-TOKEN': $('meta[name="csrf-token"]').attr('content') }
});
event.preventDefault();
// $(this).attr("disabled", "disabled");
var prdfoId = $(this).data('id');
$.ajax({
url: '{{url("orderspayonline")}}/'+encodeURI(prdfoId),
type: "POST",
cache: false,
success: function(data) {
var resultType = document.getElementById('result-type');
var resultData = document.getElementById('result-data');
function changeResult(type,data){
$("#result-type").val(type);
$("#result-data").val(JSON.stringify(data));
}
snap.pay(data, {
onSuccess: function(result){
changeResult('success', result);
console.log(result.status_message);
console.log(result);
$("#payment-form").submit();
},
onPending: function(result){
changeResult('pending', result);
console.log(result.status_message);
$("#payment-form").submit();
},
onError: function(result){
changeResult('error', result);
console.log(result.status_message);
$("#payment-form").submit();
}
});
}
});
});
</script>
The part I need to manipulate is snap.pay(data, { where results gets back.
Currently they are return in console and disappear in a sec as the result of echo $snap_token; in my controller.
I have tried to get them in hidden input, but as I mentioned I cannot get results because I can't change my try code, even I tried to get them after catch part closed, the same thing happens Error.
Question
How can I get my results in controller?
I need to update my database with that results.
Thanks.
From what I can see you have two choices.
Add another client-side AJAX call within the snap.pay function
Redirect the submission of payment-form towards your own app and then do the onward submission from the server-side.
Option 1:
function serverSubmit(d) {
$.ajaxSetup({
headers: { 'X-CSRF-TOKEN': $('meta[name="csrf-token"]').attr('content') }
});
$.ajax({
url: '{{url("newroute")}}',
type: "POST",
cache: false,
data: d
});
};
Then amend snap.pay to submit, e.g.
snap.pay(data, {
onSuccess: function(result){
changeResult('success', result);
console.log(result.status_message);
console.log(result);
serverSubmit(result); // <----------
$("#payment-form").submit();
}
...
You may want to encrypt the data if it is sensitive, and you may want to decide not to submit the form if you don't get a successful result from the serverSubmit. You may also want to submit supplemental data beyond the result, but this is a start.
Option 2
Change the <form id="payment-form... action to {{url('newroute')}}
In your controller add a new method and map a post newroute to it.
public function new_method(Request $request)
{
$data = $request->validate(...);
YourModel::create($data); //however you want to save it
$url = 'old_form_submit_URL';
$options = array(
'http' => array(
'header' => "Content-type: application/x-www-form-urlencoded\r\n",
'method' => 'POST',
'content' => http_build_query($data)
)
);
$context = stream_context_create($options);
$result = file_get_contents($url, false, $context);
if ($result === FALSE) { /* There was an error */ }
}

Call Php controller function with AJAX Codeigniter

I , im working with Ajax and Codeigniter to call function client-server
the php
public function mainViewClean() {
unset($_SESSION[$this::jsondevices]);
unset($_SESSION[$this::jsontags]);
return "Ready";
}
//route $route['cleantags'] = 'user/mainViewClean';
And ajax:
<script type="text/javascript">
$(document).ready(function(){
$("#btn_recargar").button().click(function(){
//window.location.href = "<?= base_url('home')?>";
$.ajax({
type:'POST',
url:'<?php echo base_url("cleantags"); ?>',
data:{'id':100},
success:function(data){
//window.location.href = "<?= base_url('home')?>";
alert(data);
}
});
});
});
</script>
The function excuse good , but javascript don't show any data , what im doing wrong?
Well, the ajax call reads the response from the server, and that response must be rendered as some type of readable data, such as application/json or text/html.
In order to write that data, you need to echo it from the server with PHP.
The return statement doesn't write data, it simply returns at the server level.
If you want to communicate between PHP functions, you have to use return. But if you want to output some data, you have to use echo
Client side
$.ajax({
url:'<?php echo base_url("cleantags"); ?>',
dataType: 'application/json',
success:function(response)
{
alert(response.foo);
}
})
Server Side
public function mainViewClean()
{
unset($_SESSION[$this::jsondevices]);
unset($_SESSION[$this::jsontags]);
echo json_encode( array("foo"=>"Ready"));
}
Change return into :
echo "Ready";
If you're sending an array, at server side you need to json_encode, example :
// encode array into json string format
echo json_encode( array( 'name' => 'Osman' ) );
And in Js, you have 2 options, the 1st solution is :
success : function ( data ) {
// data now is coming in this form { "name" : "osman" }
// as the string data is coming from server-side
// you must parse it back into Javascript object
var newData = JSON.parse( data );
}
And the 2nd option is, add dataType properties inside ajax properties like following :
$.ajax({
...
dataType : 'json', // with this, no need to write JSON.parse()
...
});
I'm fairly new as I only have been using AJAX , but I think your code has a few syntactical errors.
data: { id:100 } with no quotations around id.
I advise you need to look at more examples of ajax calls to fix these little errors.
You said your JS is working but not showing data?

Class Yii is not found when posting a variable with ajax

I want to get a variable from ajax to php. I'm using the Yii framework.
So my problem is, when I want to transfer a variable from ajax to a php script I get this error:
Fatal error: Class 'Yii' not found in
/var/www/vhosts/adappter.de/comamos/protected/views/store/search_area.php
on line 44 store.js:1308:13
This is how my Ajax Call looks like
var selectedCuisine = [];
$( document ).on( "click", "#cuisines", function()
{
if ( $(this).is(':checked') )
{
selectedCuisine.push($(this).val());
}
// document.getElementById('cuisine-list').style.visibility='hidden';
$.ajax({
type: "GET",
url: "../protected/views/store/search_area.php",
data: {cuisine : selectedCuisine},
success: function(response){
console.log(response);
}
});
});
and this is my php script to line 44
<?php
if (!isset($_SESSION)) { session_start(); }
$_SESSION['search_type']='';
if (isset($_GET['s'])){
$_SESSION['kr_search_address']=$_GET['s'];
$_SESSION['search_type']='kr_search_address';
}
unset($_SESSION['kr_item']);
unset($_SESSION['kr_merchant_id']);
$marker=Yii::app()->functions->getOptionAdmin('map_marker');
if (!empty($marker)){
echo CHtml::hiddenField('map_marker',$marker);
}
?>
The jQuery call is only when I click on a checkbox. So the value of the selected box gets pushed in an array. I want to return the array to the php script.
When I'm loading the site, I don't get such an error. So I don't know why this error occurs.
You need to use controller with action. Not only view file. You try to call view file from web. But it is wrong, because all code in protected directory. All requests must processing from index file. In index file connected framework. Here a small example. Controller:
class SiteController extends Controller
public function actionTest()
{
//... example
echo CHtml::button('test');
}
js:
var selectedCuisine = [];
$( document ).on( "click", "#cuisines", function()
{
if ( $(this).is(':checked') )
{
selectedCuisine.push($(this).val());
}
// document.getElementById('cuisine-list').style.visibility='hidden';
$.ajax({
type: "GET",
url: "/site/test", // url for your action
data: {cuisine : selectedCuisine},
success: function(response){
console.log(response);
}
});
});

Echo PHP message after AJAX success

I have a modal that will display when the user clicks a delete button. Once they hit the delete button I am using AJAX to subimit the form. Eveything works fine, but it is not display my success message which is set in PHP.
Here is my AJAX code:
function deleteUser(){
var id = <?php echo $userdetails['id'] ?>;
$.ajax({
type: "POST",
url: 'admin_user.php?id=' + id,
data: $('form.adminUser').serialize(),
error: function(e){
alert(e);
},
success: function () {
// This is empty because i don't know what to put here.
}
});
}
Here is the PHP code:
if ($deletion_count = deleteUsers($deletions)) {
$successes[] = lang("ACCOUNT_DELETIONS_SUCCESSFUL", array($deletion_count));
} else {
$errors[] = lang("SQL_ERROR");
}
And then I call it like this:
<div class="col-lg-12" id="resultBlock">
<?php echo resultBlock($errors,$successes); ?>
</div>
When I use AJAX it does not display the message. This works fine on other pages that does not require AJAX to submit the form.
I think you are getting confused with how AJAX works, the PHP script you call will not directly output to the page, consider the below simplified lifecycle of an AJAX request:
Main Page -> Submit Form -> Put form data into array
|
--> Send array to a script to be processed on the server
|
|----> Callback from the server script to modify DOM (or whatever you want to do)
There are many callbacks, but here lets discuss success and error
If your PHP script was not found on the server or there was any other internal error, an error callback is returned, else a success callback is fired, in jQuery you can specify a data array to be received in your callback - this contains any data echoed from your PHP script.
In your case, you should amend your PHP file to echo your arrays, this means that if a successful request is made, the $successes or $errors array is echoed back to the data parameter of your AJAX call
if ($deletion_count = deleteUsers($deletions)) {
$successes[] = lang("ACCOUNT_DELETIONS_SUCCESSFUL", array($deletion_count));
echo $successes;
} else {
$errors[] = lang("SQL_ERROR");
echo $errors;
}
You can then test you received an object by logging it to the console:
success: function(data) {
console.log(data);
}
Well, it's quite not clear what does work and what does not work, but two things are bothering me : the function for success in Ajax is empty and you have a header function making a refresh in case of success. Have you tried removing the header function ?
success: function(data) {
alert(data);
}
In case of success this would alert the data that is echoed on the php page. That's how it works.
I'm using this a lot when I'm using $.post
Your header will not do anything. You'll have to show the data on the Java script side, maybe with alert, and then afterwards redirect the user to where you want in javascript.
you need put some var in success function
success: function(data) {
alert(data);
}
then, when you read var "data" u can do anything with the text
Here is what I changed the PHP to:
if ($deletion_count = deleteUsers($deletions)) {
$successes[] = lang("ACCOUNT_DELETIONS_SUCCESSFUL", array($deletion_count));
echo resultBlock($errors,$successes);
} else {
$errors[] = lang("SQL_ERROR");
echo resultBlock($errors,$successes);
}
And the I changed the AJAX to this:
function deleteUser(){
var id = <?php echo $userdetails['id'] ?>;
$.ajax({
type: "POST",
url: 'admin_user.php?id=' + id,
data: $('form.adminUser').serialize(),
error: function(e){
alert(e);
},
success: function (data) {
result = $(data).find("#success");
$('#resultBlock').html(result);
}
});
}
Because data was loading all html I had to find exactly what I was looking for out of the HTMl so that is why I did .find.

How to export a model function from a controller to a view in Laravel 4

I am trying to display some data from my database that is dependent on some input from the user. I am using an ajax request to get the data, send it back to a function in my controller, and then export it back to my view. I would like to collect this data and display it without going to another view (I just hide the previous form and unhide the new form).
Here is the relevant code:
Javascript:
$('#submit_one').on('click', function(event) {
event.preventDefault();
if(! $(this).hasClass('faded')) {
var fbid = $("input[name='like']:checked").val();
//variable to be collected is fbid
request = $.ajax({
url: "http://crowdtest.dev:8888/fans/pick_favorite",
type: "post", success:function(data){},
data: {'fbid': fbid} ,beforeSend: function(data){
console.log(data);
}
});
to_welcome_two();
}
});
function to_welcome_two()
{
$('#welcome_one').addClass('hidden');
$('#welcome_two').removeClass('hidden');
}
Controller functions:
public function pick_favorite() {
$fbid=Input::get('fbid');
return Artist::specific_artist($fbid);
}
public function getWelcome() {
return View::make('fans.welcome')
->with('artists', Artist::artists_all())
->with('favorite_artist', Artist::favorite_artist())
->with('pick', FansController::pick_favorite());
}
Model function:
public static function specific_artist($fbid) {
$specific_artist = DB::table('artists')
->where('artists.fbid', '=', $fbid)
->get();
return $specific_artist;
}
The view is on the "welcome" page. My question is how do I display the model data in my view and make sure it is printing out the correct data from the fbid input?
I tried something like this:
#foreach($pick as $p)
<span class="artist_text">{{$p->stage_name}}</span>
<br>
<span class="artist_city">{{$p->city}}</span>
#endforeach
but this is not printing out anything. Any ideas?
i see lots of issues here.
Server side:
public function pick_favorite().... what does it do? it just returns some data.
in public function getWelcome() { , you wrote, FansController::pick_favorite(). supposing both are the same method, you are accessing a static method whilst the method is non static. you are getting an error for this but you are not seeing it because you didn't define fail().
and i don't see what the point of declaring a method which does nothing else then a model call which you can do directly.
e.g let's say i have a fooModel
public function index(){}
in controller, i can just write,
public function bar()
{
$model = new fooModel;
return View::make(array('param1'=>$model->index()));
}
or if i declare index() method in fooModel as static, then i can write,
public function bar()
{
return View::make(array('param1'=>fooModel::index()));
}
Client side:
now in your javascript,
$('#submit_one').on('click', function(event) {
event.preventDefault();
if(! $(this).hasClass('faded')) {
var fbid = $("input[name='like']:checked").val();
//variable to be collected is fbid
request = $.ajax({
url: "http://crowdtest.dev:8888/fans/pick_favorite",
type: "post", success:function(data){},
data: {'fbid': fbid} ,beforeSend: function(data){
console.log(data);
}
});
to_welcome_two();
}
});
function to_welcome_two()
{
$('#welcome_one').addClass('hidden');
$('#welcome_two').removeClass('hidden');
}
why it should print any data? you didn't asked the script to print anything. where is your .done or .success param in your code?
If you look at your console, you'l get lots of php errors, i am almost sure of.
an advice, you need to lear some basics. e.g. jquery ajax call.
a basic ajax call can be
var request = $.ajax({
url: "script.php",
type: "POST",
data: { id : menuId },
dataType: "html"
});
request.done(function( msg ) {
$( "#log" ).html( msg );
});
request.fail(function( jqXHR, textStatus ) {
alert( "Request failed: " + textStatus );
});
implement it in your code and then see what errors it throws.
Conclusion:
1st one will be (supposing rest of your codes are ok) the static error. if you want to call it as static, declare it as static. but a static function in controller? i don't see any purpose of it.
and then start the debug. your problem is both client and server side. deal one by one.

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