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);