How to pass data from javascript to php in same page

Question

I want to know how I can pass data from my script javascript to my php code to use the data into a query I tried many things but it didn't work for me So this is my script to upload files from input type: file then i get the url in downloadURL variable

 var downloadURL;
 ...
 uploadTask.on('state_changed',function(snapshot){
 },function(error){
 },function(){
 downloadURL=uploadTask.snapshot.downloadURL;
 alert(downloadURL);
 }); 

Now I want to pass downloadURL to my php so I can use it .
I also tried Ajax to do this task but it didn't work or the code that I used is false
Ajax code :

$.ajax({
 type: "POST",
 url: '', //same page
 data: downloadURL ,
 success: function(data)
 {
 //alert(data);
 }
}); 

EDIT
Php code :

<?php 
 $user=$_POST['downloadURL'];
 echo $user; 
?> 

Just a normal echo to test if data is Posted or not

• Spoody: What do you mean by didn't work? any errors?
• Amine Karimi: When I echo the result it said undefined
• Sider Topalov: You had no url for your ajax request. Give url to ajax request and will work
• Sider Topalov: Also from your php you should return any value to ajax
• Sider Topalov: Provide your php code and i will help you

Answer 1

Structure the data of your $.ajax request in a name-value pair manner.

Change this:

$.ajax({
 type: "POST",
 url: '', //same page
 data: downloadURL ,
 success: function(data)
 {
 //alert(data);
 }
}); 

To this:

$.ajax({
 type: "POST",
 data: {"downloadURL":downloadURL} ,
 success: function(data)
 {
 //alert(data);
 }
}); 

I also removed url from your $.ajax request because by default url is set to the current page.

With the above modifications, your PHP code will remain unchanged (e.g., $user=$_POST['downloadURL'];).

Answer 2

Okay change your php with that code:

if(isset($_POST['downloadURL']) {
 $response = array(
 'user' => $_POST['downloadURL']
 );
 echo json_decode($response);
 exit;
 }

Because you are making ajax request you must return json that why we parse our Array to json(object) and then in your javascript ajax request inside success function write

console.log(data);

And after data

...
data: downloadUrl,

Add this

dataType: 'json'

This mean we are telling on our ajax request that we are expecting json response