Passing empty values ​​to the database, how to solve it?

Question

PHP

Hello.

Faced such a problem. When loading the page, it passes empty values ​​to the database, after which it gives an error.

$link = mysqli_connect($host, $user, $password, $database) or die("Ошибка " . mysqli_error($link));

    // выполняем операции с базой данных
    $name = $_POST['name'];
    $lastname = $_POST['lastname'];
    $email = $_POST['email'];
    $pass = $_POST['pass'];

    $sql = "INSERT INTO `users` (`name`, `lastname`, `email`, `password`) VALUES ('$name', '$lastname', '$email', '$pass')";

    if (mysqli_query($link, $sql)) {
        echo "New record created successfully";
    } else {
        echo "Error: " . $sql . "<br>" . mysqli_error($link);
    }

    // закрываем подключение
        mysqli_close($link);

Gives this error

Notice: Undefined index: email in D:\XAMPP\htdocs\PB\registration.php on line 24

As far as I understand. It does not see the values ​​of these variables. But they are not even filled in, for some reason he immediately sends blanks to the database. They are written there, and after reloading the page it gives an error

Warning: mysqli_error() expects parameter 1 to be mysqli, null given

If I connect a send check, by type

if (isset($_POST[]))

then nothing changes. The warnings remain.

Answer 1

...
Well, everything is written the same.
You have opened the page.
Executed Since there is no post request, it does not exist, null is written to the $name variable and a notice is thrown about it. Notis does not stop the script and everything is executed further. The base includes an insert of null, which is absolutely valid. should generally cause a syntax error, because this is a meaningless entry. If check - then check specific values
$name = $_POST['name'];
if (isset($_POST[]))

if (isset($_POST['name']) && isset($_POST['lastname']) && ... и тд) {