使用php中的值到jquery

$(document).ready(function() {
  "use strict";
  var id_var = "<?php echo $id_question; ?>";

  $("#option1").click(function() {
    alert(id_var);
    $.post('updateDatabase.php', {
      id: id_var,
      option: "option1_clicked"
    });
  });

<?php 

$id = $_POST['id'];
$option = $_POST['option'];
$connection = mysqli_connect("localhost", "root", "" , "log_database");

if ($option == 'option1_clicked') {
    $queryOption1 = "UPDATE `questions` SET `option1_clicked` = `option1_clicked` + 1 WHERE id='$id'";
    $query_run_option1 = mysqli_query($connection,$queryOption1);
}
?>

When I enter in the id field of my $.post an id from my database it updates correctly. When I do an alert(id_var) to check if the variable gets the correct value, it works.

So you would think (at least I do) when I put id_var after id: it will update my database but this doesn't work.

Why?