Creating an insert statement for orders table html input form is not getting handled

I am trying to create an order form where personell of the restaurant can manually take orders from a customer. and later see what order they have made.

my statement keeps refusing to insert this data into the database. I would like to know what i am doing wrong here.,

This is my database:

MenuItem: 
    MenuItemID  int(11)         
    ItemName    varchar(255)        
    ItemPrice   double
orders:

OrderID    int(11)

MenuItemID int(11)

ReceiptID  int(11)

Res_Datum  date

Tafel_Id   int(11)

Res_ID     int(11)
receipt:

ReceiptID       int(11)

ReceiptPrice    double
reserveringen:

Reservering_Id  int(11)

Tafel_Id        int(11)

VoorNaam        varchar(255)

AchterNaam      varchar(255)

TelefoonNummer  varchar(255)

Email           varchar(255)

Res_Datum       date
tafels:

Tafel_Id        int(11)

tafel_Nummer    int(11)

Aantal_Personen int(11)

Orders are shown by orders.Tafel_Id, orders.ReceiptID, orders.MenuItemID, orders.ReceiptID, orders.Res_Datum

$sql = "SELECT O.Res_Datum, O.Res_ID, O.Tafel_Id,O.ReceiptID, SUM(MI.ItemPrice) AS TotalReceiptPrice

FROM Orders AS O INNER JOIN MenuItem AS MI ON O.MenuItemID = MI.MenuItemID

GROUP BY O.Res_Datum, O.Res_ID, O.Tafel_Id ";

$result = $conn->query($sql);

I am trying to insert this data into my database through this form, but I cannot figure out why it is not working.

Maak_bestelling.php :

<?php

$con = mysqli_connect(‘localhost’,‘root’,‘’);
if(!$con) {

echo ‘Not connected with server’;

}
if(!mysqli_select_db ($con,‘restaurant’)) {

echo ‘Database Not selected’;

}
$tablenumber = $_POST[‘tafelnummer’];

$receiptid = $_POST[‘receiptid’];

$menu_item = $_POST[‘menu_item’];

$date = $_POST[‘date’];
$sql = “INSERT INTO Orders (orders.Tafel_Id, orders.ReceiptID, orders.MenuItemID, orders.ReceiptID, orders.Res_Datum )

VALUES (‘$tablenumber’, ‘$receiptid’, ‘$menu_item’, ‘$date’)”;
if(!mysqli_query($con,$sql)){

echo ‘insert did not work’;

}else {

echo ‘Order created successfully’;

}
header(“refresh:1; url=bestelling.php”);

bestelling.php:

<form action=“/restaurant/maak_bestelling.php” method=“POST”>

<h2>Enter Order</h2>

Table Number:<br>

<input type=“text” name=“tafelnummer” value=“”><br><br>

Receipt Id:<br>

<input type=“text” name=“receiptid”   value=“”><br><br>

Menu_Item:<br>

<input type=“text” name=“menu_item”   value=“”><br><br>

Date: <br>

<input type=“date” name=“date”        value=“”><br><br>
<input type=“submit” value=“Submit”>

</form>
<h2>Pending Orders:</h2>

<?php

$servername = “localhost”;

$username = “root”;

$password = “”;

$dbname = “restaurant”;
// Create connection

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

// Check connection

if ($conn->connect_error) {

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

}
//$sql = "SELECT O.Res_Datum,O.Res_ID, O.Tafel_Id, SUM(MI.ItemPrice) AS TotalReceiptPrice FROM Orders AS O INNER JOIN MenuItem AS MI ON O.MenuItemID = MI.MenuItemID ";

$sql = "SELECT O.Res_Datum, O.Res_ID, O.Tafel_Id,O.ReceiptID, SUM(MI.ItemPrice) AS TotalReceiptPrice

FROM Orders AS O INNER JOIN MenuItem AS MI ON O.MenuItemID = MI.MenuItemID

GROUP BY O.Res_Datum, O.Res_ID, O.Tafel_Id ";

$result = $conn->query($sql);
if (mysqli_num_rows($result) > 0) {

// output data of each row

while($row = mysqli_fetch_assoc($result)) {

echo "Res_datum: “. $row[“Res_Datum”]. " ReservationID : " . $row[“Res_ID”]. " - Table_Number: " . $row[“Tafel_Id”]. " Total Order Price: " . $row[“TotalReceiptPrice”].” ". “<br>”;

}

} else {

echo “0 results”;

}
mysqli_close($conn);

?>
</div>

#php #html #forms #mysql