在PHP中捕获Ajax数据变量？

I'm trying to get the data used below to be caught in my alives.php page. Essentially, alives.php requires a variable $passcode.

How do I pass the content of data below as the variable $passcode through a POST request?

<script>
  $(document).ready(function() {

    $('#alive').click(function () {
      var data = '<?php  $row['code']; ?>';

      $.ajax({
        type:"GET",
        cache:false,
        url:"alives.php",
        data:data,    // multiple data sent using ajax
        success: function (html) {
        }
      });
      return false;
    });
  });
</script>

alives.php

<?php 
require("database.php");

$checkvote = "SELECT code FROM votes WHERE code = '$passcode'";
$updatealive = "UPDATE votes SET alive = +1 WHERE code = '$passcode'";
$addvote = "INSERT INTO votes (code, alive) VALUES ('$passcode',+1 )";

$checkvoterlt = mysqli_query($con, $checkvote); 

if(mysqli_num_rows($checkvoterlt) > 0) {
   $result = mysqli_query($con, $updatealive) or die(mysqli_error());
} else {
     $result = mysqli_query($con, $addvote) or die(mysqli_error());
}


mysqli_close($con);
?>

So much is wrong.

Problem 1: You are specifying a GET request: $.ajax({ type:"GET",. If you want it to be POST:

$.ajax({
    type:"POST",

Problem 2: Your javascript data variable should be key: value pairs like:

var data = { 'passcode' : code };

Then in PHP get the data with $_POST['passcode']

This sort of passcode should NOT be passed to the client side, as it can be easily manipulated.

Instead, store such information in a $_SESSION variable (assumes you've started your PHP with session_start), as this will keep the value safe, pass it to all pageloads where it may be needed, and be impossible to manipulate (while it is possible to hijack someone else's session, a malicious user still won't be able to actively change the value)