似乎没有执行AJAX请求

Purpose: Live PHP chat system

About the code:

The first part searches for older messages and displays them (it works). It also gets the ID of the last message that already exists.
The second part searches for new messages (their ID is greater than the one received on page loading).

Problem:

The AJAX request doen't seem to be executed since new text isn't displayed. Also, anything written after the div id="verification" doesn't work: for ex I tried displaying an alert, but without success.
I get an error in the Chrome console: Uncaught syntax error: Invalid or unexpected token on this line (and it might be beacuse of the '<div>' in the echo statement:

echo '<div>' . $row['com'] . '</div>';

Main code:

<?php
   include('config.php');
   $query = "SELECT MAX(id) FROM comments";
        $res = mysqli_query($connect,$query); 
        $rez = mysqli_fetch_assoc($res);
        $id = $rez['MAX(id)'];

        $query1 = "SELECT com FROM comments";
        $res1 = mysqli_query($connect,$query1); 
    ?>
    <!DOCTYPE html>
    <html>

    <head>
    <script src="https://ajax.googleapis.com/ajax/libs/jquery/3.4.1/jquery.min.js"></script>
    <style>
    .verification{
      overflow-y:scroll;
      height:300px;
      width:100px;
    }
    </style>

    </head>

    <body>

    <div id="verification">
    <?php
    while($row = mysqli_fetch_assoc($res1))
             { 
                 echo '<div>' . $row['com'] . '</div>';
             }   
    ?>
    </div>

    <script>

    setInterval(function () {
    $.ajax({
       type: "GET",
       url: "getmsg.php?id=<?php echo $id; ?>",
       dataType: "json",
       success: function(data) {
         for (i = 0; i < data.length; ++i) {
           var msg = data[i][0];
           $("#verification").append('<div class="left spotmsg"><div class="message">
           <p class="txt">'+msg+'</p></div></div>');
           }
        }
      });
    }, 1000);

    </script>

    </body>

    </html>

getmsg.php:

<?php
include('config.php');
$id1 = $_GET['id'];
$query = "SELECT com FROM comments WHERE id > $id1";
$res = mysqli_query($connect,$query);

while ($row = mysqli_fetch_all($res)) {
header('Content-Type: application/json');
ob_end_clean();
echo json_encode($row);
}
?>