I have been trying to fix this issue I'm having but I am not able to figure out how. This is my current code:
if (isset($_POST['username'])){
$username = stripslashes($_POST['username']);
$username = mysqli_real_escape_string($con,$username);
$email = stripslashes($_POST['email']);
$email = mysqli_real_escape_string($con,$email);
$password = stripslashes($_POST['password']);
$password = mysqli_real_escape_string($con, $password);
$password = md5($password);
$geboorte_datum = $_POST['geboorte_datum'];
$naam = $_POST['naam'];
$adres = $_POST['adres'];
$postcode = $_POST['postcode'];
$woonplaats = $_POST['woonplaats'];
$query = "
INSERT INTO
`users` (id, username, password, email, geboortedatum, medewerker, volledige_naam, volledige_adres, postcode, woonplaats)
VALUES
(NULL, '$username', '$password', '$email', '$geboorte_datum', '0', '$naam', '$adres', '$postcode', '$woonplaats')
";
$result = mysqli_query($con,$query);
if($result){
echo "
<div class='form'>
<h3>Je bent succesvol geregistreerd</h3>
<br/>Klik hier om <a href='inloggen.php'>in te loggen</a>
</div>
";
} else {
echo "
<div class='form'>
<h3>Fout opgetreden</h3>
<br/>Klik hier om <a href='registreren.php'>opnieuw te proberen</a>
</div>
";
}
}else{
?>
This unfortunately gives me 4 errors that are all the same:
mysqli_real_escape_string()
expects parameter 1 to bemysqli
, null given in....
Would any of you guys be able to figure this out? Much appreciated!