Uncaught Error: Syntax error, unrecognized expression. What to do?

Question

AJAX

I want to implement sending and instantly displaying a comment via ajax.
I pass on the necessary information. I return back the finished markup of the new comment.

Here is the full error code:


jquery code:

function sendComment() {
   const sendButton = $('.post__write-comment-sbmt');

   sendButton.each(function () {
      $(this).click(function (e) {
         e.preventDefault();
         const clicked = $(this);

         //ПЕРЕМЕННЫЕ ДЛЯ ПЕРЕДАЧИ ПО AJAX
         const message = clicked.siblings('.post__write-comment-write').val()
         const postID = clicked.closest('.post').attr('id');
         const commentsCount = clicked.closest('.post').find('.post__comment-count');
         //___________________________________________________________________________________

         $.ajax({
            type: 'POST',
            url: 'php-scripts/likes-commentshandler.php',
            data: {
               sendComment: '',
               postID: postID,
               message: message,
               commentsCount: commentsCount.text(),
            },
            dataType: 'html',
            success: function (data) {

               const comItems = clicked.siblings('.post__com-items');

               comItems.append($(data));

               commentsCount.text(commentsCount + 1);
            }
         });
      });
   });
}

PHP code:

function sendComment()
{
   require __DIR__ . '/../#DATABASE.php';
   if (isset($_POST['sendComment'])) {
      //---------------ПЕРЕМЕННЫЕ-----------------------
      $postID = $_POST['postID'];
      $userID = $_SESSION['user']['id'];
      $message = htmlspecialchars($_POST['message']);
      $commentsCount = $_POST['commentsCount'];

      $name = $_SESSION['user']['name'];
      $surname = $_SESSION['user']['surname'];
      //________________________________________________

      //---------------mysql-запрос---------------------
      $query = $db->prepare(
         "INSERT INTO `comments` (`postid`, `userid`, `message`, `id`) VALUES ('$postID', '$userID', :message, NULL)"
      );
      $param = [
         'message' => $message
      ];
      $query->execute($param);
      //_____________________________________________________

      //-------------------Ответ ajax------------------------

      echo 'div class="post__com-item"><div class="post__com-header"><div class="post__com-author"><a href="another-page.php?uid=' . $userID . '" class="post__com-author-link"><img src="img/ava.jpg" alt="" class="post__com-ava"><span class="post__com-nickname">' . $name . ' ' . $surname . '</span></a></div></div><div class="post__com-body"><p class="post__com-message">' . $message . '</p></div></div>';

      //______________________________________________________
   }
}

Answer 1

Missing "<" opening div tag