Question

我正在尝试创建一个订单表单，餐厅的人员可以手动从客户那里接受订单。然后看他们做了什么订单。

我的语句一直拒绝将这些数据插入数据库。我想知道我在这里做错了。，

这是我的数据库：

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)

按订单显示订单。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);

我正在尝试通过此表单将这些数据插入数据库中，但是我无法弄清为什么它不起作用。

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>

Answer 1

在插入语句中，您重复了ReceipID，因此必须像这样删除它：

$sql = "INSERT INTO Orders (orders.Tafel_Id, orders.ReceiptID, orders.MenuItemID,  orders.Res_Datum )
VALUES ('$tablenumber', '$receiptid', '$menu_item', '$date')";

没有为订单表html输入表单创建插入语句

1 个答案: