Inserting data by PHP and MySQL - php

Question

It's working, but when I add the data in to my database, the data will be twice. I don't know if my syntax is wrong or my code is wrong.
Here's the structure:
//if submit is clicked
$checkin = $_POST['text_checkin'];
while ($row = mysqli_fetch_array($reservation)) {
if (isset($_POST['submitBtn'])) {
if ($row['reservefrom'] == $checkin) {
echo "Same Date";
return;
}
else
{
$lastname = $_POST['text_lastname'];
$firstname = $_POST['text_firstname'];
$address = $_POST['text_address'];
$tnumber = $_POST['text_tnumber'];
$cnumber = $_POST['text_cnumber'];
$email = $_POST['text_email'];
$checkin = $_POST['text_checkin'];
$checkout = $_POST['text_checkout'];
$room = $_POST['text_room'];
$tour = $_POST['text_tour'];
$guest = $_POST['text_guest'];
$query = "INSERT INTO reservation
(lastname, firstname, homeaddress,
telephonenumber, cellphonenumber, email,
reservefrom, reserveto, room, tour,
guestnumber)
values ('$lastname', '$firstname', '$address',
'$tnumber', '$cnumber', '$email', '$checkin',
'$checkout', '$room', '$tour', '$guest')";
mysqli_query($db, $query);
echo "Data Submitted!";
}
}
}

Answer 1

You're getting multiple inserts because you are looping for each record in $reservations. You should first look into why you are getting multiple records if you expected just a single record reservation.
That aside, alter your code by replacing your while loop with:
if(isset($_POST['submitBtn']) && $row = mysqli_fetch_array($reservation)){
if($row['reservefrom'] == $checkin) die("Same Date");
$lastname = $_POST['text_lastname'];
$firstname = $_POST['text_firstname'];
// ... other values, then execute your query
}else{
// either submitBtn was not posted or no result were found in $reservation
}
I noticed also that you use return in your code, but the code doesn't seem to be within a function so that's confusing. If it is within a function, it's probably a bad idea to echo from within unless the function is specifically meant to send data directly to the browser.