I have a question regarding ajax, I'm quite new to it so not sure of the best procedure. Anyways I have incorporated ajax in to my CodeIgniter app but I have the possibility of 2 different responses and I'm not quite sure how to deal with this in my ajax.
Instead of appending my result to a div can I not just refresh the url?
In my controller I have form validation, when it is false I want to refresh to display my errors, but if it returns true I want to show the new page, if that makes sense?
view
$.post("<?php echo base_url(); ?>part-one",
$("form").serialize(),
function(result){
// if errors show this
$("#error").html(result);
// if there are no errors how do I check the response to refresh the page to a new url?
}, "html"
);
controller
if($this->form_validation->run() == FALSE){
$data['content'] = "part_one";
$this->load->view('template', $data);
} else {
$data['content'] = "part_two";
$this->load->view('template', $data);
}
If you want to reload the page you can do;
$.ajax({
url: url,
type: 'post',
data: data,
success: function(data) {
location.reload();
}
});
If i understand you right then this code may help you ...
$.post("<?php echo base_url(); ?>part-one",
$("form").serialize(),
function(result){
// if errors show this
$("#error").html(result);
if(result =='false condition response')
{
window.location.reload()
}
if(result == 'True Conditon response')
{
window.location.href('URL of the page')
}
// if there are no errors how do I check the response to refresh the page to a new url?
}, "html"
);
Related
I wanted to submit a for using ajax call in laravel 5.
In view i wrote something like
$("#updateSubmit").on('submit',function(e){
e.preventDefault();
var csrfToken = $('meta[name="csrf-token"]').attr("content");
$.ajax({
method:'POST',
url: '/account/updateForm',
//dataType: 'json',
data: {accountId:'1111', _token: '{{csrf_token()}}'},
success: function( data )
{
alert(data)
return false;
}
},
error: function(error ){
alert("There is some error");
}
});
and on controller side
public function update(Request $data )
{
return Response()->json(['success' => true],200);
}
while in route for post method
Route::post('account/updateForm', 'AccountController#update')->name('account/updateForm');
its working till ajax. on Submission of ajax it goes to controller action.
but it does not retrun back as ajax comes back in normal form submisson.
it just go to controller and stops there with {"success":true} line.
I want ajax to come back to view form so that I can perform different dependent actions.
Do you mean that when you submit your form, you just have a white page with {"success": true} ?
If that's the case, maybe the error is on your javascript.
Maybe your jQuery selector is wrong, or maybe your js isn't compiled ?
EDIT
Ok, so I can login fine but when I enter false info I'm redirected to the login page, what I need is to stay on the same page and show the error message e.preventDefault(); doesn't seem to work.
$(function() {
$("#login-form").submit(function(e) {
$('.fail').hide();
$.ajax({
url:"/login",
type: "post",
data: $(this).serialize(),
error:function(){
$('.fail').show();
e.preventDefault();
},
success: function(){
document.location = '/';
}
});
return false;
});
});
Your not actually doing anything with the form, ill try commenting your code to talk you through whats happening.
I'm guessing your using PHP server side for this code.
In PHP you want to check the user credentials and then tell the browser. If the login was successful send back "y", and if it failed "n".
<script type="text/javascript">
$(function() {
$("#login-form").submit(function() {
$('.fail').hide();
$.ajax({
url:"/login",
type: "post",
data: $(this).serialize(),
error:function(){
$('.fail').show();
},
success: function(data) {
if (data == "y") {
//Login was successful. Redirect statement here?
} else {
//Failed login message here
}
}
});
return false;
});
});
</script>
Edit
Try adding this in your success function. Please let me know what you get for a successful login and a failed login.
success: function(data) {
console.log(data);
}
Edit 2
This is because your ajax call is successful, just that the login failed. This is why your success handler is called.
To sort this you'll need to see what is being returned from the server, is it nothing? In which case try:
success: function(data) {
if (data == "") {
e.preventDefault();
} else {
//Login successful, redirect user.
}
}
You should add:
success: function(data){
/* Validation data here, if authentication answer is correct */
if(data == 'ok')
document.location = '/';
else
/* show error here */
}
the success occur when URL is found and accessible.
and error occur when URL is not found.
if you want check the callback MSG you must print it in the URL page & check data in success.
like this:
<script type="text/javascript">
$(function() {
$("#login-form").submit(function() {
$('.fail').hide();
$.ajax({
url:"/login",
type: "post",
data: $(this).serialize(),
success:function(data){
if(data=="true"){
alert("success");
}else{
alert("faild")
}
}
});
return false;
});
});
</script>
in login.php
<?php
//check if for true...
echo "true";
//and if it is not true...
echo "false";
?>
the data parameter in success is a string which get back the html content of "/login"
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 : $form_data,
success: function(data) {
swal({
title: "Updated!",
text: "Your entity was updated successfully",
type: "success"
},
function(){
location.reload(true);
});
}
});
});
I am creating a simple wordpress plugin and trying to use AJAX, but I always get 0 in ajax response.
<script type="text/javascript" >
jQuery(document).ready(function($) {
var data = {
action: 'my_action',
whatever: '1234'
};
jQuery.post("http://localhost/taichi/wp-admin/admin-ajax.php", data, function(response) {
alert(response);
});
});
</script>
<?php
add_action('wp_ajax_my_action', 'my_action_callback');
add_action( 'wp_ajax_nopriv_my_action', 'my_action_callback' );
function my_action_callback() {
echo "test";
die();
}
what am I doing wrong?
You have to put the add_action at the complete bottom of your file or else it won't find the callback function
Try to change :
jQuery.post("http://localhost/taichi/wp-admin/admin-ajax.php", data, function(response)
To :
jQuery.post(ajaxurl, data, function(response)
And check if it is working on the admin side first. It should work fine.
Error Return Values
If the AJAX request fails when the request url is wp-admin/admin-ajax.php, it will return either -1 or 0 depending on the reason it failed.
Read this
Edit
admin-ajax always return default '0' as output.so while you alerting response you will 0 only.using die() in callback function will terminate that.
Had the same problem, it turned out that my callback was inside a php file which was only included to my "Theme Options" page.
To check if the function is able to trigger trougth admin-ajax.php try to add var_dump(function_exists("your_callback_name")); to the bottom of the wp-admin/admin-ajax.php (before die( '0' );) and then have a look to your ajax output.
Try the following code in your plugin file. or in function.php
jQuery(document).ready(function($){
var ajaxURL = 'http://localhost/taichi/wp-admin/admin-ajax.php';
var dataString = 'action=mnd_news';
$.ajax({
type: "POST",
url: ajaxURL,
data: dataString,
cache: false,
success: function(response){
if(response != 'error') {
alert(response);
}
}
});
});
add_action('wp_ajax_mnd_news', 'get_mnd_ajax');
add_action( 'wp_ajax_nopriv_mnd_news', 'get_mnd_ajax' );
function get_mnd_ajax() {
echo "test";
die();
}
I'm tryin to validate an input field with an ajax call to a cakephp controller
My Ajax is:
$("#UserAlphaCode").change(function () {
$.ajax({
type: "post",
url: '<?php echo $this->webroot ?>' + "/alpha_users/checkCode",
data: ({code : $(this).val()}),
dataType: "json",
success: function(data){
alert (data);
},
error: function(data){
alert("epic fail");
}
});
});
My controller code
function checkCode() {
Configure::write('debug', 0);
$this->autoRender = false;
$codePassed = $this->params['form']['code'];
$isCodeValid = $this->find('count',array('conditions'=> array('AlphaUser.code' => $codePassed)));
if ($isCodeValid == 0){
$codeResponse = false;
} else {
$codeResponse = true;
}
echo json_encode ($codeResponse);
}
I'm pretty sure I'm using $this->params wrong here to access the data sent from the ajax request. What should I be doing instead?
Try something like:
$codePassed = $_POST['code']
you might also try putting:
$this->log($codePassed,LOG_DEBUG);
somewhere in there and examine the output in tmp/logs/debug.log
Using firebug will help debug the transport.
Don't know why it would be returning null, but I normally use $this->data to fetch form data.
And did you try debug($this->params)? If you don't have a non-AJAX form to test the request from, use Firebug or Wireshark to see what is being return by the server for the debug() call—since it will break jQuery's AJAX handler by not being in JSON.