How to declare a PHP variable and pass it into AJAX url?

Question

When i call /comment.php it returns data that i've requested. where post_id is fixed.

/comment.php

 "ajax":
{
 "url": "http://localhost/fb-callback.php?post_id=12345_67890",
   "type" : "GET",
"dataSrc":  function (response) { 
                // console.log(response);
                            if(response.whaterver == 0){
                               //DO YOUR THING HERE
                            }
                            //return back the response
                            return response; 
                        },

Instead of fixed id (12345_67890), i wanna declare a php variable there, something like this ::-

<?php
$uid = $_GET['uid'];
?>

 "ajax":
{
 "url": "http://localhost/fb-callback.php?post_id="+"&uid",
"type" : "GET",
 ................

so that when i'll call url /comment.php?uid=12345_67890** it'll return results. But it seems it doesn't work that way. HOW CAN I DO THAT Or is there other way to make this call ?

Here is the Full code ::: https://pastebin.com/iy3htv63

"url": "http://localhost/fb-callback.php?post_id=<?= $uid ?>", see also here: stackoverflow.com/questions/2150238/… — user5734311
– user5734311, Commented May 8, 2019 at 6:12

Syed mohamed aladeen · Accepted Answer · 2019-05-08 06:10:21Z

3

you can echo it on the ajax url. try this

 <?php
  $uid = $_GET['uid'];
?>

 "ajax":
{
 "url": "http://localhost/fb-callback.php?post_id=<?php echo $uid; ?>",
 "type" : "GET",
 ................

answered May 8, 2019 at 6:10

Syed mohamed aladeen

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2 Comments

Anon Over a year ago

thanks, thats really helps me a lot... appreciate it :)

Pure Rhymer Organization Over a year ago

This is a bad coding practice.

SAVe · Accepted Answer · 2019-05-08 06:17:16Z

0

Try This

<?php
$uid = $_GET['uid'];
?>
var uid = <?php echo $uid; ?>;
 "ajax":
{
 "url": "http://localhost/fb-callback.php?post_id="+uid,
"type" : "GET",
 ................

edited May 8, 2019 at 6:17

answered May 8, 2019 at 6:11

SAVe

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