Undefined variable:

So where in your code are you defining $linkup?

attaching the page edit-linkup.php
edit-linkup.php

This error means that within your code, there is a variable or constant which is not set. But you may be trying to use that variable.

Check if you have select the 'details' field from database in your select query statement.

I don't see $linkup set anywhere in that additional page you provided. It is referenced several times but not actually assigned a value anywhere.

I did the change...it comes up blank...definitely data in the details field of tbl_answers_therapist

FYI - David's suggestion only circumvents the error message and shows a blank instead of the error. It does not resolve the underlying root cause, which is that $linkup is not set in your code. You need to define $linkup.

My guess is that maybe processForm.php is supposed to define $linkup. Can you share that file's code?

Thank you. So how do I define $linkup and where?

Here you go..

<?php
ini_set('display_errors', 1);
ini_set('display_startup_errors', 1);
error_reporting(E_ALL);

//var_dump($client_id);
//var_dump($client);

$msg = "";
$msg_class = "";

$conn = mysqli_connect(DB_SERVER,DB_USER,DB_PASS,DB_NAME);
// Check connection
if (mysqli_connect_errno())
{
 echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
  if (isset($_POST['save_profile'])) {
    // for the database
    //$profileImageName = time() . '-' . $_FILES["profileImage"]["name"];
   $profileImageName = $_FILES["profileImage"]["name"];  
    // For image upload
    $target_dir = "../AVATAR/images/";
    $target_file = $target_dir . basename($profileImageName);
    // VALIDATION
    // validate image size. Size is calculated in Bytes
    if($_FILES['profileImage']['size'] > 2000) {
      $msg_class = "alert-danger";
     $msg = "Image size should not be greated than 2MB";
      
    }
    // check if file exists
    if(file_exists($target_file)) {
      $msg_class = "alert-danger";
     $msg = "Avatar already exists";
      
    }
    // Upload image only if no errors
    if (empty($error)) {
      if(move_uploaded_file($_FILES["profileImage"]["tmp_name"], $target_file)) {
        $sql = "UPDATE tbl_therapist SET therapist_avatar='$profileImageName' WHERE therapist_id = {$therapist_id} LIMIT 1";
      
      if(mysqli_query($conn, $sql)){
          $msg_class = "alert-success";
        $msg = "Avatar uploaded and saved";
          
        } else {
          $msg_class = "alert-danger";
        $msg = "There was an error in the database";
          
        }
      } else {
        $msg = "alert-danger";
      $error = "There was an error uploading your avatar";
      }
    }
  }
?>

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That's your application's logic, none of us can tell you that. We can only guess.

I mean, based on usage, I would assume that you should probably have some code that queries the desired row from tbl_answers_therapist as an object and then assigns that value to a variable called $linkup but that's just a guess.

Again, my guess is that that code would be somewhere in processForm.php since it's included at the top but again, just an educated guess. I would have to know your architecture and application pretty well before I could tell you for sure (and that level of help would warrant a separate, paid consulting gig).

Hi, No, I did not write the code a friend did.  Playing with it. Will update. Thank you.