我需要 table 方面的帮助。它们没有组织/整齐并且非常困惑 ->
表格必须看起来整洁、易于理解并且不会像这样困惑 -> (这就是表格的样子)
这是我迄今为止所取得的成就(参见:OrderID 2)-> http://i.imgur.com/fj06EGB.png
下面是代码
<table>
<tr>
<th>Customer Name</th>
<th>Customer Contact</th>
<th>Customer Email</th>
<th>Order ID</th>
<th>Order Date</th>
<th>Menu Name</th>
<th>Price</th>
<th>Quantity</th>
<th>Total Amount</th>
<th>Action</th>
</tr>
<?php
$result = $mysqli->query("SELECT * FROM `order`");
while($obj = mysqli_fetch_assoc($result)) {
$orderDate = $obj['OrderDate'];
$orderId = $obj['OrderID'];
$totalAmount = $obj['OrderTotal'];
$paymentStatus = $obj['PaymentStatus'];
$customerId = $obj['CustomerID'];
$res = $mysqli->query("SELECT CustomerName, CustomerContactNo, CustomerEmail FROM customer WHERE CustomerID=$customerId");
if($row = mysqli_fetch_assoc($res)) {
$customerName = $row['CustomerName'];
$customerContactNo = $row['CustomerContactNo'];
$email = $row['CustomerEmail'];
}
$result1 = $mysqli->query("SELECT * FROM ordermenu WHERE OrderID = $orderId");
while($obj1 = mysqli_fetch_assoc($result1)) {
$menuId = $obj1['MenuID'];
$menuQty = $obj1['menuQty'];
$result2 = $mysqli->query("SELECT * FROM menu WHERE MenuID = $menuId");
$obj2 = mysqli_fetch_assoc($result2);
$name = $obj2['MenuName'];
$price = $obj2['MenuPrice'];
?>
<tr>
<td><?php echo $customerName;?></td>
<td><?php echo $customerContactNo;?></td>
<td><?php echo $email;?></td>
<td><?php echo $orderId;?></td>
<td><?php echo $orderDate;?></td>
<td><?php echo $name;?></td>
<td>$<?php echo $price;?></td>
<td><?php echo $menuQty;?></td>
<td>$<?php echo $totalAmount;?></td>
<td><a href="update_order.php?id=<?php echo $orderId;?>" color="green">Update</a></td>
</tr>
<?php } ?>
<?php } ?>
</table>
请大家帮忙。
最佳答案
- 首先从数据库获取客户
对于每个客户,按 customerID 与 ordermenu(在 ordermenu.OrderID = order.OrderID)连接并与 menu(在 ordermenu.MenuID = menu.MenuID)连接的方式获取订单
<?php $customerQuery = $mysqli->query("SELECT CustomerID, CustomerName, CustomerContactNo, CustomerEmail FROM customer;"); while($customer = mysqli_fetch_assoc($customerQuery)) { $customerId = $customer['CustomerID']; $orderQuery = $mysqli->query("SELECT * FROM `order` o LEFT JOIN ordermenu om ON o.OrderID = om.OrderID LEFT JOIN menu m ON m.MenuID = om.MenuID WHERE o.CustomerID = $customerId"); while($order = mysqli_fetch_assoc($orderQuery)) { // do something with your customer and order record // show in table for example } }
关于php - 无组织的表,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/28311728/