从jquery文件上传插件uploadhandler文件获取响应

Question

你好朋友我正在使用jquery文件上传插件来上传文件。当我上传任何文件然后它成功移动到文件夹并将文件信息存入数据库。 但现在我想在成功插入数据库后回复。

我也总是有一个错误：Error SyntaxError: Unexpected token < 错误的图像看起来像这样：

这是我正在执行数据库插入查询并将图像移动到文件夹中的函数。

protected function handle_file_upload($uploaded_file, $name, $size, $type, $error,
            $index = null, $content_range = null) {
            GLOBAL $con;
            $time=time();
            $folder_path="../../uploads/media/";
            $rand=rand();
    $user_id=$_SESSION['id'];
        $file = new stdClass();
        $file->name = $this->get_file_name($uploaded_file, $name, $size, $type, $error,
            $index, $content_range);
        $file->size = $this->fix_integer_overflow(intval($size));
        $file->type = $type;
        if ($this->validate($uploaded_file, $file, $error, $index)) {
            $this->handle_form_data($file, $index);
            $upload_dir = $this->get_upload_path();
            if (!is_dir($upload_dir)) {
                mkdir($upload_dir, $this->options['mkdir_mode'], true);
            }
            $file_path = $this->get_upload_path($file->name);
            $append_file = $content_range && is_file($file_path) &&
                $file->size > $this->get_file_size($file_path);
            if ($uploaded_file && is_uploaded_file($uploaded_file)) {
                // multipart/formdata uploads (POST method uploads)
                if ($append_file) {
                    file_put_contents(
                        $file_path,
                        fopen($uploaded_file, 'r'),
                        FILE_APPEND
                    );
                } else {
                    //move_uploaded_file($uploaded_file, $file_path);
                    ////
                    $filename = date('YmdHis',$rand).$_SESSION['id'].$name;
                    $file_size=$size;
                    $file_size=$file_size/1000;
                    move_uploaded_file($uploaded_file,$folder_path.$filename);
                    mysqli_query($con,"insert into gallery(login_id,pic,timestamp,file_size,status)values('$user_id','$filename','$time','$file_size','1')");
                     $gallery_id=mysqli_insert_id($con);

                    ////
                }
            } else {
                // Non-multipart uploads (PUT method support)
                file_put_contents(
                    $file_path,
                    fopen('php://input', 'r'),
                    $append_file ? FILE_APPEND : 0
                );
            }
            $file_size = $this->get_file_size($file_path, $append_file);
            if ($file_size === $file->size) {
                $file->url = $this->get_download_url($file->name);
                if ($this->is_valid_image_file($file_path)) {
                    $this->handle_image_file($file_path, $file);
                }
            } else {
                $file->size = $file_size;
                if (!$content_range && $this->options['discard_aborted_uploads']) {
                    unlink($file_path);
                    $file->error = $this->get_error_message('abort');
                }
            }
            $this->set_additional_file_properties($file);
        }
        return $file;
    }

现在从上面的函数我想要响应gallery_id 如何在成功上传文件后删除此错误并获得响应。

你是天才的人请帮助我，并询问是否有不清楚的事情。