Not receiving data from ajax to php - javascript

I want to send data from javascript to another php page where I want to display it. I found that I need to use Ajax to pass the data to php so I tried myself.
My file where is the javascript:
$('#button').on('click', function () {
$.jstree.reference('#albero').select_all();
var selectedElmsIds = [];
var selectedElmsIds = $('#albero').jstree("get_selected", true);
var i = 0;
$.each(selectedElmsIds, function() {
var nomenodo = $('#albero').jstree('get_selected', true)[i].text;
//var idnodo = selectedElmsIds.push(this.id);
var livellonodo = $('#albero').jstree('get_selected', true)[i].parents.length;
//console.log("ID nodo: " + selectedElmsIds.push(this.id) + " Nome nodo: " + $('#albero').jstree('get_selected', true)[i].text);
//console.log("Livello: " + $('#albero').jstree('get_selected', true)[i].parents.length);
i++;
$.ajax({
type: "POST",
data: { 'namenodo': nomenodo,
'levelnodo': livellonodo
},
success: function(data)
{
$("#content").html(data);
}
});
});
});
I want to send the data to another php page which consists of:
<?php echo $_POST["namenodo"]; ?>
But when I try to go to the page there's no data displayed.

This is a very basic mistake I think every beginner (including me) does while posting a data using ajax to another php page.
Your ajax code is actually posting the data to lamiadownline.php (if you are using the variables correctly) but you can't get that data by simply using echo.
Ajax post method post data to your php page (lamiadownline.php) but when you want to echo the same data on the receiver page (lamiadownline.php), you are actually reloading the lamiadownline.php page again which makes the $_POST["namenodo"] value null.
Hope this will help.

First of all you won't be able to see what you have post by browsing to that page.
Secondly, is this
<?php echo $_POST["namenodo"]; ?>
in the current page?
Otherwise, specify the url
$.ajax({
url: "lamiadownline.php",
type: "POST",
data: { 'namenodo': nomenodo,
'levelnodo': livellonodo},
success: function(data) {
$("#content").html(data);
}
});

//try this
$.ajax({
type: "POST",
url:"Your_php_page.php"
data: { namenodo: nomenodo levelnodo: livellonodo},
success: function(data)
{
$("#content").html(data);
}
});

Related

How to GET javascript data in PHP file without page reload

I am building my best attempt at a twitter clone and have run into a bit of a problem. I want to be able to click on a post and, without a page refresh, display that post in the overlay of the page (as you would on a twitter feed to look at replies, etc.).
In script.js, I check for a click and try to change the url.
$('body').on("click", ".chirp", function(){
var uid = $_GET['id'];
var pid = $(this).attr("id");
var pidSplit = pid.split("chirp");
var messageID = pidSplit[1];
var obj = {foo: "status"};
$('.chirpOverlay').addClass("active");
window.history.pushState(obj, "Status", "profile.php?id="+uid+"&status="+pid);
});
The javascript works as intended...but as I will soon find out, the victory is short-lived.
In profile.php, I attempt to GET the status id from the URL parameter.
<?php
$status_id = $_GET['status'];
$sql = $db->query("SELECT * FROM chirps WHERE id='$status_id'");
if (mysqli_num_rows($sql) > 0) {
$c = $sql->fetch_object();
}
?>
This doesn't work because, as I've learned, using 'window.history.pushState' only changes the url- but doesn't load the page. Thus the $_GET statement fails. I need a way to get the id of the post I click on into profile.php without a page refresh. Even if it means taking a different approach (instead of using a URL parameter).
PS: I tried to do an XMLHttpRequest as well- to no avail. :(
Thanks in advance!
$('body').on("click", ".chirp", function(){
var uid = $_GET['id'];
var pid = $(this).attr("id");
var pidSplit = pid.split("chirp");
var messageID = pidSplit[1];
var obj = {foo: "status"};
$('.chirpOverlay').addClass("active");
$.ajax({
url: "profile.php?id="+uid+"&status="+pid,
type: "GET",
data: obj,
dataType: "html",
success: function(data){
console.log(data);
}
});
});
You need to just get something up and going that works and then you can add more to it as you figure things out. This should give you a good starting place.
Here are your two files. Make sure they are both in the same directory.
You will need to make sure you have a jquery version loaded. Put this on whatever page you are calling the script.js from.
<script src="https://ajax.googleapis.com/ajax/libs/jquery/3.3.1/jquery.min.js"></script>
script.js
$(document).ready(function(){
$('body').click(function(){
var id; //define your id.
var pid; //define your pid.
var datastring = 'id=' + uid + '&status=' + pid;
console.log(datastring);
$.ajax({
url: 'profile.php',
type: 'POST',
cache: false,
data: datastring,
dataType: 'json',
success: function(data){
console.log('Made it to the success function: ' + data);
if (data) {
//It works, do something.
console.log(data);
} else{
//It does not work, do something.
console.log('Your ajax failed to get the info from your php page. Keep troubleshooting');
}
}
});
});
});
profile.php
<?php
/*
$status_id = $_POST['status']; //This needs to be sanitized and validated.
$sql = $db->query("SELECT * FROM chirps WHERE id='$status_id'"); //This is unsafe sql practice.
if (mysqli_num_rows($sql) > 0) {
$c = $sql->fetch_object();
}
echo json_encode($c); //This is what gets sent back to script.js
*/
echo 'You made it to your php page.';
?>
A few things:
You can not call any php variable from within your js. var uid = $_GET['id']; does not work.
Any value that you pass to the php page needs to be validated to make sure it is a legitimate value.
Your SQL query is prone to sql injections. Please read up on how to parameterize your queries. Good Mysqli Practices
I have finally found a AJAX-based solution to my problem.
I created a new php file called "chirp_open_ref.php" and added this ajax to script.js:
var datastring = 'status=' + messageID;
$.ajax({
url: "chirp_open_ref.php",
type: "POST",
data: datastring,
cache: false,
dataType: "text",
success: function(data){
$('.chirp-container').html(data);
}
});
Inside of 'chirp_open_ref.php':
<?php
require 'core.inc.php';
if (isset($_POST['status']) && isset($_SESSION['user_id'])){
$chirp_id = $_POST['status'];
$c = "";
$sql = $db->query("SELECT * FROM chirps WHERE id='$chirp_id'");
if (mysqli_num_rows($sql) > 0){
$c = $sql->fetch_object();
}
include'chirp.inc.php';
}
?>
'chirp.inc.php' is simply a template for the layout/structure of each post.
This works like a charm, but I am always open to any criticism of how I am performing this. Thanks for all the help guys!

AJAX call returns undefined in Header / form-data

I am trying to get the contents from some autogenerated divs (with php) and put the contents in a php file for further processing. The reason for that is I have counters that count the number of clicks in each div. Now, I ran into a problem. When I echo back the data from the php file, the call is made, but I get undefined in the form-data section of the headers, and NULL if I do var_dump($_POST). I am almost certain I am doing something wrong with the AJAX call. I am inexperienced to say the least in AJAX or Javascript. Any ideas? The code is pasted below. Thanks for any help / ideas.
The AJAX:
$(document).ready(function(e) {
$("form[ajax=true]").submit(function(e) {
e.preventDefault();
var form_data = $(this).find(".test");
var form_url = $(this).attr("action");
var form_method = $(this).attr("method").toUpperCase();
$.ajax({
url: form_url,
type: form_method,
data: form_data,
cache: false,
success: function(returnhtml){
$("#resultcart").html(returnhtml);
}
});
});
});
The PHP is a simple echo. Please advise.
Suppose you have a div
<div id="send_me">
<div class="sub-item">Hello, please send me via ajax</div>
<span class="sub-item">Hello, please send me also via ajax</span>
</div>
Make AJAX request like
$.ajax({
url: 'get_sorted_content.php',
type: 'POST', // GET is default
data: {
yourData: $('#send_me').html()
// in PHP, use $_POST['yourData']
},
success: function(msg) {
alert('Data returned from PHP: ' + msg);
},
error: function(msg) {
alert('AJAX request failed!' + msg);
}
});
Now in PHP, you can access this data passed in the following manner
<?php
// get_sorted_content.php
if(!empty($_POST['yourdata']))
echo 'data received!';
else
echo 'no data received!';
?>
It's sorted. Thanks to everyone. The problem was I didn't respect the pattern parent -> child of the divs. All I needed to do was to wrap everything in another div. I really didn't know this was happening because I was echoing HTML code from PHP.

Ajax change php variable

I've got this variable $type and I want it to be month or year.
It should be changed by pressing a div.
I've tried creating an onclick event with an ajax call.
The ajax call and the variable are in the same script (index.php)
Inside the onclick function:
var curr_class = $(this).attr('class');
$.ajax({
type: "POST",
url: "index.php",
data: {
type: curr_class
},
dataType: 'text',
success: function(data) {
// Test what is returned from the server
alert(data);
}
});
But the alert returns the whole html page.
When I console.log the data (create a var data = { type:curr_class }) and console.log *that data* it returnstype = month` (which is correct)
while I just want it to return month or year
So on top of the page I can call
if(empty($_POST['type'])){
$type = 'month';
} else {
$type = $_POST['type'];
}
and change the PHP variable so I can use it in the rest of my script.
But how can I accomplish this?
With kind regards,
as you are sending request to the same page so as a result full page is return .You will have to send it to another page and from that page return the type variable
if(empty($_POST['type'])){
$type = 'month';
} else {
$type = $_POST['type'];
echo $type;
keep this code in separate file and make an ajax call to that page
//Try This It's Work
Get Value
Get Value
$(".btn-my").click(function(){
var curr_class = $(this).data('title');
$.ajax({
type: "POST",
url: "index.php",
data: {
type: curr_class
},
dataType: 'text',
success: function(data) {
// Test what is returned from the server
alert(data);
}
});
});

redirect page using ajax & jquery (php)

i want to redirect from page a to page profile and in between there is a post session on them. in this case let's say the data is variable $name in string. so far my code is like this on page a
jQuery("#result").on("click",function(e){
var $clicked = $(e.target);
var $name = $clicked.find('.name').html();
var decoded = $("<div/>").html($name).text();
$('#searchid').val(decoded);
//the ajax script
$.ajax({
type: 'POST',
url: 'b.php',
data: 'result='+$name,
success: function() {
window.location.href = "profile.php"; // replace
}
});
});
and on page b the code is:
<?php echo $_POST['result']?>
the outcome should be the value from result in which determined on page a.
but so there is an error message saying unidentified index. so where am i doing wrong?
Could it be, that your data parameter is wrong?
I have my ajax calls as folowing:
jQuery.ajax({
type: "POST",
url: "b.php",
data: {
result: $name
},
success: function() {
window.location.href = "profile.php"; // replace
}
});
It is a new request after the redirect. In order to access the result you need to sotre it in som kind of session or pass it again.
You can pass it like this, then it will be in $_GET
success: function(data) {
window.location.href = "profile.php?result="+data; // replace
}

Make page refresh if successful login with ajax/jquery

Ajax code
$("#login_btn").click(function() {
var url = "core/login.php";
$.ajax({
type: "POST",
url: url,
data: $("#sr").serialize(),
success: function(data) {
var responseData = jQuery.parseJSON(data);
$('.oops').html('<div class="error">'+responseData.oops+'</div>');
alert(responseData.good);
if(responseData.good === 1) {
alert(good)
location.reload();
}
}
});
});
Php code if everything passes
else {
$_SESSION['id'] = $login;
$good = 1;
exit();
}
How come it's not freshing the page with location.reload? Should I be using .done .try?
Dont refresh, the whole idea of using ajax is that you dont need to.
You can set your sessions in the ajax php script and pull new content based on the change in session.
That being said, you are missing a ';' at
alert(good);

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