Question

我已经做了一张表格，以便员工可以在餐厅的桌子上接单。当人员为客户进行预订时，数据库中会生成一个预订ID。

我想做的是在Orders.Res_ID上加入reserveringen.Reservering_Id，这样我下订单时就不必手动输入Res_ID。我只想输入日期和表格，然后知道属于哪个预订。

这是我正在使用的语句，它可以正常工作，但是即使我的数据库中有保留，reservation_id仍为空。

$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  ";

这给了我他们在特定日期订购的总金额。我唯一不知道的是什么reservering.Reservering_Id坐在那张桌子上。

这是查询的结果：我手动输入了此价格，但总价格确实可以计算和工作。

Res_datum: 2019-01-21 ReservationID : 5 - Table_Number: 3 BonNummer: 3 Total Order Price: 12 
Res_datum: 2019-01-22 ReservationID : 0 - Table_Number: 5 BonNummer: 2 Total Order Price: 12

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

任何帮助将不胜感激。

这是我正在使用的一些代码：

SELECT * FROM `orders` INNER JOIN reserveringen on Orders.Res_ID = reserveringen.Reservering_Id

现在发生的是，我获得了reserveringen表中的所有信息，并且它会自动填充所有内容。 但是我确实需要手动输入reservering.Reservering_Id，现在它不再计算出总收货价格。

这是我的代码：

<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);
}

//MY Query i am trying to run
// $sql = "SELECT O.Res_Datum, R.Reservering_Id, O.Tafel_Id,O.ReceiptID, SUM(MI.ItemPrice) AS TotalReceiptPrice
// FROM Orders, Reserveringen AS O INNER JOIN MenuItem AS MI ON O.MenuItemID = MI.MenuItemID
// AS O LEFT JOIN Reservering_Id AS R on O.Reservering_Id = R.reservering_Id
// GROUP BY O.Res_Datum, O.Res_ID, O.Tafel_Id  ";

//The statement I get with an Empty reservation_ID
$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  ";

SELECT * FROM `orders` JOIN reserveringen on Orders.Res_ID = reserveringen.Reservering_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: " . $row["Tafel_Id"]. " BonNummer:  ".$row["ReceiptID"]."  Total Order Price: " . $row["TotalReceiptPrice"]." ". "<br>";
    }
} else {
    echo "0 results";
}

mysqli_close($conn);
?>


</div>

除非我以表单

0 个答案: