无法插入数据库

Question

我按照教程制作了一个登录系统，并且学到了很多东西。但是现在我正在尝试提供视频共享服务，而我尝试插入数据库中的所有信息实际上都没有。我想我已经正确编码了，但是似乎没有用。

我正在使用xampp并使用PHP和Apache。我尝试更改一些代码，甚至重新创建了我要向其中插入数据的表。

// All the database code
require 'dbh.inc.php';

if(isset($_POST["video-submit"])) { // Where the user presses the submit

    // Getting the variables
    $file = $_FILES['video'];
    $filename = $file['name'];
    $fileexp = explode('.', $filename);
    $fileext = strtolower(end($fileexp));
    $title = $_POST['title'];
    $desc = $_POST['desc'];
    $author = $_SESSION['userid'];
    $sql = "SELECT videoid FROM videos WHERE videoid = ?"; // Finds the videoid
    $stmt = mysqli_stmt_init($conn);
    if (!mysqli_stmt_prepare($stmt, $sql)) { // Error checking
        header("location: ../upload.php?upload=fail&reason=sqlerror&title=".$title."&desc=".$desc);
        exit();
    } else {
        mysqli_stmt_bind_param($stmt, "s", $videoid);
        mysqli_stmt_execute($stmt);
        mysqli_stmt_store_result($stmt);
        $result = mysqli_stmt_num_rows($stmt);
        if ($result > 0) { // Checking if $videoid already exists
            header("location: ../upload.php?upload=fail&reason=sqlerror&title=".$title."&desc=".$desc);
            exit();
        } else {
            $sql = "INSERT INTO videos (videoid, video, title, description, author, views, likes, dislikes) VALUES (?, ?, ?, ?, ?, ?, ?, ?)";
            $stmt = mysqli_stmt_init($conn);

            if (!mysqli_stmt_prepare($stmt, $sql)) {
                header("location: ../upload.php?upload=fail&reason=sqlerror&title=".$title."&desc=".$desc);
                exit();
            } else {
                $views = $likes = $dislikes = 0;
                $video = $videoid.".".$fileext;
                $filedest = "../videos/".$videoid;
                move_uploaded_file($filetmp, $filedest);

                mysqli_stmt_bind_param($stmt, "sssssiii", $videoid, $video, $title, $desc, $author, $views, $likes, $dislikes);
                mysqli_stmt_execute($stmt);

                header('location: /watch.php?v='.$video);
                exit();
            }
        }
    }
    mysqli_stmt_close($stmt);
    mysqli_close($conn);
}

我希望将数据插入数据库，但是在代码运行后进行检查时，没有任何迹象。该注册代码可以完美运行。请告诉我这是否是一个愚蠢的错误。