Newbie seeking help with malfunctioning code

I’m a beginner who likes to start with PHP as a hobby. I’m writing a questionnaire in html/css and want the answers to be saved in a server I made in MySQL. When a question isn’t filled in I want a message to appear “Gelieve alle vragen te beantwoorden.” (translation = Please fill in all questions), and when all questions are filled in they just need to be saved in the server.

But I keep getting errors. Can someone please explain where I’m making mistakes and how to solve the errors, please?

This is the code from the PHP file:

<?php

include 'Connection_db.php';

/* Haalt alle gegevens uit het formulier */

if (isset($_POST ['Submit'])){
    
    if (!isset($_POST['Email'], $_POST['Voornaam'], $_POST['Achternaam'], $_POST['Geslacht'], $_POST['Postcode'], $_POST['A1'], $_POST['A2'], $_POST['A3'], $_POST['A4'], $_POST['A5'], $_POST['A6'], $_POST['A7'], $_POST['A8'], $_POST['A9'])) {
        echo "Gelieve alle vragen te beantwoorden.";
    } else {    
    $Email = $_POST['Email'];
    $Voornaam = $_POST['Voornaam'];
    $Achternaam = $_POST['Achternaam'];
    $Geslacht = $_POST['Geslacht'];
    $Postcode = $_POST['Postcode'];
    $A1 = $_POST['A1'];
    $A2 = $_POST['A2'];
    $A3 = $_POST['A3'];
    $A4 = $_POST['A4'];
    $A5 = $_POST['A5'];
    $A6 = $_POST['A6'];
    $A7 = $_POST['A7'];
    $A8 = $_POST['A8'];
    $A9 = $_POST['A9'];
    }
    
    $connection = mysqli_connect ('localhost', 'root', '', 'wolf');     /* 'Servernaam', 'gebruikersnaam', 'wachtwoord', 'databasenaam' */
    if (!$connection) {
        die ('connection failed');}
    
    $query = "INSERT INTO klantentevredenheid_2019 (Email, Voornaam, Achternaam, Geslacht, Postcode, A1, A2, A3, A4, A5, A6, A7, A8) "; /* Naam tabel (kolommen) */
    $query .= "VALUES ('$Email','$Voornaam','$Achternaam','$Geslacht','$Postcode','$A1','$A2','$A3','$A4','$A5','$A6','$A7','$A8','$A9')";        /*ingevulde waarde */
    
    $result = mysqli_query ($connection, $query);
        if (!$result) {
            die ('query failed' . mysqli_error());
        }
        
}

?>

And these are the errors I keep getting:

Gelieve alle vragen te beantwoorden.connection completed
Notice: Undefined variable: Email in C:\xampp\htdocs\Wolf\Likert_form_php.php on line 34

Notice: Undefined variable: Voornaam in C:\xampp\htdocs\Wolf\Likert_form_php.php on line 34

Notice: Undefined variable: Achternaam in C:\xampp\htdocs\Wolf\Likert_form_php.php on line 34

Notice: Undefined variable: Geslacht in C:\xampp\htdocs\Wolf\Likert_form_php.php on line 34

Notice: Undefined variable: Postcode in C:\xampp\htdocs\Wolf\Likert_form_php.php on line 34

Notice: Undefined variable: A1 in C:\xampp\htdocs\Wolf\Likert_form_php.php on line 34

Notice: Undefined variable: A2 in C:\xampp\htdocs\Wolf\Likert_form_php.php on line 34

Notice: Undefined variable: A3 in C:\xampp\htdocs\Wolf\Likert_form_php.php on line 34

Notice: Undefined variable: A4 in C:\xampp\htdocs\Wolf\Likert_form_php.php on line 34

Notice: Undefined variable: A5 in C:\xampp\htdocs\Wolf\Likert_form_php.php on line 34

Notice: Undefined variable: A6 in C:\xampp\htdocs\Wolf\Likert_form_php.php on line 34

Notice: Undefined variable: A7 in C:\xampp\htdocs\Wolf\Likert_form_php.php on line 34

Notice: Undefined variable: A8 in C:\xampp\htdocs\Wolf\Likert_form_php.php on line 34

Notice: Undefined variable: A9 in C:\xampp\htdocs\Wolf\Likert_form_php.php on line 34

Warning: mysqli_error() expects exactly 1 parameter, 0 given in C:\xampp\htdocs\Wolf\Likert_form_php.php on line 38
query failed

Thanks for the help!

I fixed this for you, please use english naming next time!

  <?php

    include 'Connection_db.php';

    /* Haalt alle gegevens uit het formulier */

    function out()
    {
        echo 'Gelieve alle vragen te beantwoorden.';
    }

    function hasData()
    {
        $aVars = [
            'Email', 'Voornaam', 'Achternaam', 'Geslacht', 'Postcode'
        ];

        foreach ($aVars as $sVar) {
            if (!isset($sVar)) {
                out();
                return false;
            }
        }

        for ($i = 1; $i < 10; $i++) {
            if (!isset($_POST["A{$i}"])) {
                out();
                return false;
            }
        }

        return true;
    }

    if (isset($_POST ['Submit'])) {

        if (hasData()) {
            $Email = $_POST['Email'];
            $Voornaam = $_POST['Voornaam'];
            $Achternaam = $_POST['Achternaam'];
            $Geslacht = $_POST['Geslacht'];
            $Postcode = $_POST['Postcode'];
            $A1 = $_POST['A1'];
            $A2 = $_POST['A2'];
            $A3 = $_POST['A3'];
            $A4 = $_POST['A4'];
            $A5 = $_POST['A5'];
            $A6 = $_POST['A6'];
            $A7 = $_POST['A7'];
            $A8 = $_POST['A8'];
            $A9 = $_POST['A9'];
        }

        $connection = mysqli_connect('localhost', 'root', '', 'wolf');     /* 'Servernaam', 'gebruikersnaam', 'wachtwoord', 'databasenaam' */
        if (!$connection) {
            die ('connection failed');
        }

        $query = 'INSERT INTO klantentevredenheid_2019 (Email, Voornaam, Achternaam, Geslacht, Postcode, A1, A2, A3, A4, A5, A6, A7, A8) '; /* Naam tabel (kolommen) */
        $query .= "VALUES ('$Email','$Voornaam','$Achternaam','$Geslacht','$Postcode','$A1','$A2','$A3','$A4','$A5','$A6','$A7','$A8','$A9')";        /*ingevulde waarde */

        $result = mysqli_query($connection, $query);
        if (!$result) {
            die ('query failed' . mysqli_error($connection));
        }

    }

The problem you are having is with the
if (!isset($_POST[‘Email’]… line

Try something like this:
if(! isset($_POST[‘Email’]) || ! isset($_POST[‘Voornaam’] … and so on))

Don’t separate the condition in the if statement with a coma!