Passing Back multiple PHP variables after a ajax Form Call - php

Question

I am passing a file to a php file via ajax and i am returning only 1 $ variable using die($var) in the php file after a sucsessfull run...
the problem i am now facing is passing more than 1 variable back to the ajax sucess function . i have tried using json encode but it has failed to work. im thinking maybe to do with the ajax being form data.
im hoping there is a simple way top pass multiple varibles back to the sucess function.
Any help is greatly appreciated
var form_data = new FormData(); // Creating object of FormData class
form_data.append("image", file , newimagesrc) // Appending parameter named file with properties of file_field to form_data
form_data.append("oldimagesrc", oldimagesrc) // to re-write over with new image
form_data.append("email", email)
form_data.append("imagext", fileNameSub)
$.ajax({
url: "UploadProfileImage.php",
type: "POST",
data: form_data,
processData: false,
contentType: false,
success: function(newimagesrc){
//how do i pass back from php these variables
var out1=out1;
var out2=out2;
alert(out1 , out2);
//help appreciated
var newimagesrc = newimagesrc;
//alert(newimagesrc); alert recieved message
imagename=input.files[0].name;
$('#imageupdate').css('color','green');
$('#imageupdate').text(newimagesrc);
var refreshimage = "Profileimagerefresh.php?avatar="+newimagesrc+"&email="+email;
$('#imagerefresh').load(refreshimage);
}//success 1 messagereturn1
});//ajax1
PHP FILE ('UploadProfileImage.php')
if(file_exists($oldimagelocation) && is_readable($oldimagelocation)){
$new=$rnd.$accountname.".".$extension;
if ($stat->execute(array("$new","$email"))){
unlink($oldimagelocation);
die($oldimagesrc); //HERE I PASS 1 $ BACK - I NEED TO RETURN MORE
exit();
}
else{
die("Failed replace image with image and rename");
exit();
}
}

Answer 1

Using JSON encode is the best choice. I would recommend something like this:
if (file_exists($oldimagelocation) && is_readable($oldimagelocation)) {
$new = $rnd.$accountname.".".$extension;
if ($stat->execute([$new, $email])) {
unlink($oldimagelocation);
echo json_encode([
'out1' => $oldimagelocation,
'out2' => $oldimagesrc,
], JSON_FORCE_OBJECT);
} else {
die("Failed replace image with image and rename");
}
}
Then in JS just parse the response as JSON
$.ajax({
url: "UploadProfileImage.php",
type: "POST",
data: form_data,
processData: false,
contentType: false,
success: function(newimagesrc){
let jsonObj = JSON.parse(newimagesrc);
console.log(jsonObj.out1);
}
});

Answer 2

I have encountered a similar issue recently and solved this using JSON.
With PHP you can put the variables into an array and then use json_encode: and then return that onto the webpage.
Now, using your jQuery, you can use $.parseJSON which then makes it an array in jQuery.
Here's an example:
PHP:
die(json_encode(array('Hello', 'world!')));
jQuery:
$.ajax({
type: 'POST',
url: 'test.php',
}).done(function(result){
var array = $.parseJSON(result);
alert(array[0]);
});