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Hei folkens. Takk for gode svar. Skal sjekke ut litt mer.

Her er da både html, ajax og PHP:

Kode

<form action="" method="post" id="delete-user">
    <label for="delete">Delete my user from the records</label> <br>
    <input type="email" name="delete" placeholder="Your email"><br>

    <label for="skey">SecretKey:</label>
    <input type="text" name="skey" placeholder="Your secret key">


    <button type="submit">Delete me!</button>
    <p id="delete-message"></p>
</form>

Kode

$("#delete-user").submit(function(e) {
    e.preventDefault();

    $.ajax({
        url: "****/delete_form_data.php",
        method: "post",
        data: $("form").serialize(),
        dataType: "text",
        success: function(strMessage) {
            $("#delete-message").text(strMessage);
            $("#delete-user")[0].reset();
        }
    });
});

Kode

$conn = new mysqli($servername, $username, $password, $dbname);

if ($conn->connect_error) {
    die("Connection failed: " . $conn->connect_error);
}


$delete = $_POST['delete'];
$skey = $_POST['skey'];

$query = mysqli_query($conn, "SELECT * FROM form WHERE email='$delete'");

$sql = "DELETE FROM form WHERE secretKey='$skey' AND email='$delete'";

if(mysqli_num_rows($query) < 0) {
	echo "The email: " . $_POST['delete'] . " does not exist in our records";
}
elseif(!mysqli_query($conn, $sql)) {
	echo "Could not delete data!";
}
else {
	echo "You have deleted your information from our site. We hope you will come back!";
}


mysqli_close($conn);