Why is my jquery/ajax post not registering in my php code? - javascript

So I am trying to build a single page application using PHP and AJAX but I am having some issues with my navigation.
I have a file controller.php that controls the page to be displayed. The code that handles a post request is as follows
EDIT: Forgot to mention that echoing $_POST['page'] shows nothing
if(!isset($_POST['page'])){
include('Pages/landing_page.php');
echo "empty";
}
else{
echo $_POST['page'];
if($_POST['page'] == 'landing_page'){
include('Pages/landing_page.php');
}
if($_POST['page'] == 'forum'){
include('Pages/forum.php');
echo "forum hit";
}
}
The post request is generated with $.ajax function as follows
$(document).ready(function(){
$("#home_link").on('click', ()=>{
alert("Working");
$.ajax({
type: "POST",
url: "controller.php",
data: "page=landing_page"
});
});
$('#forum_link').on('click', ()=>{
$.ajax({
type: "POST",
url: "controller.php",
data: "page=forum",
success: function(){
alert("Callback Works");
}
});
});
});
I know the jquery click is working because I get the original "Working" alert as well as the callback alert if I click on the forum link.
I am using the Netbeans built in PHP development server but I have also tested it on a proper apache2 server that is well configured for PHP and AJAX.
Any help would be appreciated, thanks!

As I understand the question, this should work:
PHP: (all the same)
if(!isset($_POST['page'])){
include('Pages/landing_page.php');
echo "empty";
}
else{
echo $_POST['page'];
if($_POST['page'] == 'landing_page'){
include('Pages/landing_page.php');
}
if($_POST['page'] == 'forum'){
include('Pages/forum.php');
echo "forum hit";
}
}
JS:
$(document).ready(function(){
$("#home_link").on('click', ()=>{
alert("Working");
$.ajax({
type: "POST",
url: "controller.php",
data: "page=landing_page"
});
});
$('#forum_link').on('click', ()=>{
$.ajax({
type: "POST",
url: "controller.php",
data: "page=forum",
success: function(data){
$("#container").html(data);
alert("Callback Works");
}
});
});
});
You need to take the data returned from the function and put it somewhere, presumably into a container element.
If this isn't what you are trying to do, please comment.

Related

Yii2 Ajax Submission not working

Iam new to Yii2 and Ajax
I want to add multiple job for a work ,for that I pass id to WorkJobs Controller
This is my code for ajax submission
<?php
$this->registerJs(
'$("body").on("beforeSubmit", "form#w1", function() {
var form = $(this);
if (form.find(".has-error").length) {
return false;
}
$.ajax({
var jobid = "<?php echo $id;?>";
url: form.attr("work-jobs/create&id="+jobid),
type: "post",
data: form.serialize(),
success: function(errors) {
alert("sdfsdf");
// How to update form with error messages?
}
});
return false;
});'
);
?>
But it's not working ,I don't know what's wrong in my code ,please help ...........
change your code like below
<?php
$url=Yii::$app->urlManager->createUrl(['work-jobs/create','id'=>$id]);
$this->registerJs(
'$("body").on("beforeSubmit", "form#w1", function() {
var form = $(this);
if (form.find(".has-error").length) {
return false;
}
$.ajax({
url: "$url",
type: "post",
data: form.serialize(),
success: function(errors) {
alert("sdfsdf");
// How to update form with error messages?
}
});
return false;
});'
);
?>
Building off jithin's answer, make the following changes to your $.ajax() call
Make sure your URL is in quotes. It is a common mistake to forget to quote the URL when interspersing it with PHP. [jithin]
Unlike jithin's answer, you should do the following
instead of responding to the beforeSubmit event, handle the submit event. This would allow the Yii clientsoide validations do their job
the ajax.success callback takes data as the argument; not error, there's the ajax.failure callback for errors
Try using createAbsoluteUrl() in url like this:
url: "<?php echo Yii::app()->createAbsoluteUrl(\"work-jobs/create&id=\")"+jobid

codeigniter or PHP - how to go to a URL after a specific AJAX POST submission

I am successfully inserting data into my database in codeigniter via a an ajax post from javascript:
//JAVASCRIPT:
$.ajax({
type: "POST",
url: submissionURL,
data: submissionString,
failure: function(errMsg) {
console.error("error:",errMsg);
},
success: function(data){
$('body').append(data); //MH - want to avoid this
}
});
//PHP:
public function respond(){
$this->load->model('scenarios_model');
$responseID = $this->scenarios_model->insert_response();
//redirect('/pages/view/name/$responseID') //MH - not working, so I have to do this
$redirectURL = base_url() . 'pages/view/name/' . $responseID;
echo "<script>window.location = '$redirectURL'</script>";
}
But the problem is that I can't get codeigniter's redirect function to work, nor can I get PHP's header location method to work, as mentioned here:
Redirect to specified URL on PHP script completion?
either - I'm guessing this is because the headers are already sent? So as you can see, in order to get this to work, I have to echo out a script tag and dynamically insert it into the DOM, which seems janky. How do I do this properly?
Maybe you can 'return' the url in respond function and use it in js
PHP :
public function respond(){
// code
$redirectURL = base_url() . 'pages/view/name/' . $responseID;
return json_encode(['url' => $redirectURL]);
}
JS :
$.ajax({
type: "POST",
url: submissionURL,
data: submissionString,
dataType: 'JSON',
failure: function(errMsg) {
console.error("error:",errMsg);
},
success: function(data){
window.location = data.url
}
});
you have to concatenate the variable. That's all.
redirect('controller_name/function_name/parameter/'.$redirectURL);

pass js variable to php using ajax on the same page

this is my html code:
<form id="form" action="javascript:void(0)">
<input type="submit" id="submit-reg" value="Register" class="submit button" onclick="showtemplate('anniversary')" style='font-family: georgia;font-size: 23px;font-weight: normal;color:white;margin-top:-3px;text-decoration: none;background-color: rgba(0, 0, 0, 0.53);'>
</form>
this is my javascript code:
function showtemplate(temp)
{
$.ajax({
type: "POST",
url: 'ajax.php',
data: "section="+temp ,
success: function(data)
{
alert(data);
}
});
}
this is my ajax.php file:
<?php
$ajax=$_POST['section'];
echo $ajax;
?>
The above html and javascript code is included in a file named slider.php. In my index file i have included this slider.php file and slider.php is inside slider folder. So basically index.php and slider.php are not inside the same folder.
Javascript code alerts the data properly. But in my php code (ajax.php file) the value of $_POST['section'] is empty. What is the problem with my code. I tried googling everything and tried a few codes but it still doesn't work. Please help me out
Try this instead:
$.ajax({
type: "POST",
url: 'ajax.php',
data: { 'section': temp},
success: function(data)
{
alert(data);
}
});
It is quite possible that your server does not understand the string you have constructed ( "section="+temp ). When using ajax I prefer sending objects since for an object to be valid it requires a certain format.
EDIT1:
Try this and let me know if it doesn't work either:
$.post('ajax.php', {'section': temp}, function(data}{
alert(data);
});
Add jquery plugin(jQuery library) ,then only ajax call works
for eg
<script src="http://ajax.googleapis.com/ajax/libs/jquery/1.11.2/jquery.min.js"></script>
check your input data , whether it contain '&,/' charators,then use encodeURIComponent()
for Ajax Call Eg:
var gym_name = encodeURIComponent($("#gym_name").val());
$.ajax({
type: "POST",
url: "abc.php",
data:string_array,
success: function(msg) {
console.log(msg);
}
});
In abc.php
<?php
$id = $_POST['gym_name'];
echo "The id is ".id;
?>
Give a try to the following (although I cannot work out why #Grimbode's answer is not working):
$("#submit-reg").on( "click", function() {
$.ajax({
type: "POST",
url: 'ajax.php',
data: {'section': 'anniversary'},
success: function(data)
{
alert(data);
}
});
});
Note: I don't know what your underlying code is doing. However, I would suggest not using HTML element properties to handle events for numerous reasons, but separate the JS/event handling appropriately (separate js file(s) (recommended) or inside <script> tags). Read more...

Ajax call (before/success) doesn't work inside php file

I have a form that when clicked the submit button makes a call via ajax. This ajax is inside a php file because I need to fill in some variables with data from the database. But I can not use the calls before / success. They just do not work, I've done tests trying to return some data, using alert, console.log and nothing happens. Interestingly, if the ajax is isolated within a file js they work. Some can help me please?
File:
<?php
$var = 'abc';
?>
<script type="text/javascript">
$(document).ready(function() {
$('#buy-button').click(function (e){
var abc = '<?php echo $var; ?>';
$.ajax({
type: 'POST',
data: $('#buy-form').serialize(),
url: './ajax/buy_form.php',
dataType: 'json',
before: function(data){
console.log('ok');
},
success: function(data){
},
});
});
});
</script>
HTML:
<form id="buy-form">
<div class="regular large gray">
<div class="content buy-form">
/* some code here */
<div class="item div-button">
<button id="buy-button" class="button anim" type="submit">Comprar</button>
</div>
</div>
</div>
</form>
----
EDIT
----
Problem solved! The error was in the before ajax. The correct term is beforeSend and not before. Thank you all for help.
You said it was a submit button and you do not cancel the default action so it will submit the form back. You need to stop that from happening.
$('#buy-button').click(function (e){
e.preventDefault();
/* rest of code */
Now to figure out why it is not calling success
$.ajax({
type: 'POST',
data: $('#buy-form').serialize(),
url: './ajax/buy_form.php',
dataType: 'json',
before: function(data){
console.log('ok');
},
success: function(data){
},
error : function() { console.log(arguments); } /* debug why */
});
});
My guess is what you are returning from the server is not valid JSON and it is throwing a parse error.
Try this
<script type="text/javascript">
$(document).ready(function() {
$('#buy-button').click(function (e){
e.preventDefault();
var abc = '<?php echo $var; ?>';
$.ajax({
type: 'POST',
data: $('#buy-form').serialize(),
url: './ajax/buy_form.php',
dataType: 'json',
beforeSend: function(data){
console.log('ok');
},
success:function(data){
}
});
});
});
And make sure that your php file return response

Javascript passing variable issue not returning

Hi All I have the following code to pass a JS variable using AJAX as seen below:
function buttonCallback(obj){
var id = $(obj).attr('id');
$.ajax({
type: "POST",
url: "/project/main/passid",
data: { 'id': id },
success: function(msg){
window.alert(msg);
}
});
}
if I put an alert box in I can see the object id is successfully getting grabbed. however if in php I want to simply return the variable - I am getting a null has anyone got any ideas:
heres my PHP function (I am using Codeigniter):
public function passid(){
$courseId = $this->input->post('id');
echo $courseId;
}
EDIT: the success alert box appears - but appears blank and that is my issue I am hoping to see the ID
1. Can you make sure id equals something by doing this:
function buttonCallback(obj){
var id = $(obj).attr('id');
alert( id ); // What does it alert?
$.ajax({
type: "POST",
url: "/project/main/passid",
dataType: "json",
data: { 'id': id },
success: function(msg){
window.alert(msg.id);
}
});
}
2. Your javascript looks good... Can you try this instead to see if it works:
JS
function buttonCallback(obj){
var id = $(obj).attr('id');
$.ajax({
type: "POST",
url: "/project/main/passid",
dataType: "json",
data: { 'id': id },
success: function(msg){
window.alert(msg.id);
}
});
}
PHP
public function passid(){
$courseId = $this->input->post('id');
$response = array( "id" => $courseId );
header('Content-Type: application/json');
echo json_encode( $response );
}
3. If this does not work, can you try and rename from id to something like poopoo, just to make sure id is not taken and being weird?
4. Can you check what the network response is of your ajax request - Go to developer toolbar and goto network section, make sure you hit the record/play button. then send your request off. When you see your request come up in the network list, check the "details" of it and goto response.

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