问输入值在单击一次后不会立即更改- PHP、AJAX、JQuery

EN

<?php
$host="localhost";
$username="root";
$password="";
$databasename="admin_example";

$connect = mysqli_connect($host, $username, $password, $databasename);
$select = mysqli_query($connect, "SELECT * FROM rating");
$data = mysqli_fetch_assoc($select);

if( $connect -> connect_error){
   echo ("connection failed" . $connection->connect_error);
}

function query($query){
    global $connect;
  
    $result = mysqli_query($connect, $query);
    $rows = [];
    while($row = mysqli_fetch_array($result)){
        $rows[] = $row;
  
    }
    return $rows;
  }

if(isset($_POST['like'])){
    global $connect, $select;

    $query = mysqli_query($connect, "UPDATE rating SET total_votes=total_votes+1,likes=likes+1");
    //which should be executed after the update
    $select = mysqli_query($connect, "SELECT * FROM rating");
    foreach($select as $row){
        $likes = $row["likes"];

        echo $likes;
        exit();
      }
}

?>

<link rel="stylesheet" type="text/css" href="rating_style.css">

<input type="button" id="like" onclick="like();" value="Likes: <?php echo $data["likes"]; ?>">


<script type="text/javascript" src="http://code.jquery.com/jquery-1.7.1.min.js"></script>

<script>

function like(){
    $.ajax({
          type: 'POST',
          url: 'index.php',
          data: {
            like:"likes"
          },
          success: function (response) {
            $("#like").prop('value', 'Likes: ' + response);
          }
    });
    }


</script>

<?php



?>