【发布时间】：2017-08-03 00:46:02
【问题描述】：
我需要一些关于如何在我的数据库中编辑我的结构的想法我正在为 1 个外卖客户构建一个快餐订购网站，此时用户将那里的项目添加到订单表中
这是桌子：http://prntscr.com/g3o3kz
user_id 列是为订购 id 的用户存储的，然后 item id 是他们想要的项目 takeawayid 是外卖的 id 并且状态是他们是否订购食物，其自动设置为等待然后带走可以将其更新为接受或拒绝。
现在我遇到的问题是用户在上表中添加了 5 个项目作为外卖 id 1，用户 ID 将有 5 行，外卖 id 和项目 ID（有 5 个项目）
现在我进入了最近的订单页面，我打印了每个外卖的所有内容，就像这样
但是如果他们明天订购，它将添加另外 5 件商品，它会显示在同一个框中，我将如何对每个订单进行分组？所以订单 1 有 xyz 项目，然后下一次订单 order2 有 xyz 等等。因为目前数据库的布局方式如果有意义的话，它将都是 order1
显示订单框
<div class="container" style="margin-top: 40px;">
<div class="row">
<div class="col-xs-12">
<?php
echo $_SESSION['username'] ;
foreach ($ordersByUsers as $userId => $item) {
$username = $item['username'];
$ordersPrice = $item['ordersPrice'];
$orders = $item['orders'];
?>
<div class="row">
<div class="col-xs-12 col-md-10" style="background-color: #FAFAF8; border: 1px solid #ddd; padding: 20px; margin-bottom: 20px;">
<div class="row" style="padding: 20px;">
<div class="col-xs-12 col-sm-3">
<div class="row">
<a href="#">
<img src="http://santetotal.com/wp-content/uploads/2014/05/default-user.png" class="img-responsive" alt="Restaurant logo" height="92" width="102">
</a>
</div>
<div class="row" style="padding-top: 10px;">
<a href="#" style="text-transform: uppercase;">
<a href="#" style="text-transform: uppercase;"> </a>
<?php echo $username; echo "<br> </br>";
?>
<?php
/////////// get status of order
$statement8 = $db->prepare("SELECT * FROM users WHERE username = ? ");
$statement8->execute(array($_SESSION['username']));
$count8 = $statement8->fetch();
$username233 = mysql_real_escape_string($count8['id']);
$username2312312 = strip_tags($username233);
$statement85 = $db->prepare("SELECT * FROM orders WHERE user_id = ? AND takeawayid = ? ");
$statement85->execute(array($username2312312,$_SESSION['userid']));
$count84 = $statement85->fetch();
if($count84 ['status']== "waiting") {
?>
<form action="dashboard.php" method="post">
<input type="hidden" name="accept" value="<?php echo $count84['id']?>">
<input type="image" name="submit" src="http://business.fsdfsdfsdf.com/beta/accept.png" border="0" alt="Submit" />
</p>
</form>
<form action="dashboard.php" method="post">
<input type="hidden" name="deny" value="<?php echo $count84['id']?>">
<input type="image" name="submit" src="http://business.sdfsdfdf.com/beta/deny.png" border="0" alt="Submit" />
</p>
</form>
<?php
}elseif ($count84 ['status']== "Accepted") {
echo "<img src='http://business.sfdfsdfsdf.com/beta/accepted.png' alt='error'>";
} else {
echo "<img src='http://business.sdfdsfsdfsdf.com/beta/declined.png' alt='error'>" ;
}
?>
</a>
</div>
</div>
<div class="col-xs-12 col-sm-9">
<?php
foreach ($orders as $order) {
$orderId = $order['orderId'];
$itemname = $order['itemname'];
$price = $order['price'];
$date = $order['date'];
$address = $order['address'];
?>
<div class="row">
<div class="col-xs-12" style="background-color: #FFF; border: 1px solid #ddd; padding: 20px; margin-bottom: 20px;">
<h4>
<a href="profile.html" style="color: orange;">
<?php echo $itemname; ?>
</a>
</h4>
<div>
<div>
<span>
<?php echo $date; ?>
</span>
</div>
<div class="ratings">
<span>5 STARS</span>
</div>
</div>
</div>
Address: <b> <?php echo $address; ?></b>
</div>
<?php
}
?>
</div>
</div>
</div>
<div class="col-xs-12 col-md-2" style="padding: 20px; padding-top: 40px; font-size: 20px; color: #666; text-transform: uppercase;">
Total: <?php echo $ordersPrice; ?>
</div>
</div>
<?php
}
?>
查询
function fetchPaidOrdersByTakeaway($connection, $takeawayId, $paid) {
if (!isset($takeawayId)) {
throw new Exception('Takeaway ID not provided!');
}
if (!isset($paid)) {
throw new Exception('Paid not provided!');
}
// Sql statement.
$sql = 'SELECT
ord.id AS orderId,
usr.id AS userId,
usr.username,
ord.itemname,
ord.address,
ord.price,
ord.date
FROM orders AS ord
LEFT JOIN users AS usr ON usr.id = ord.user_id
WHERE
user_id = :takeawayid
AND paid = :paid
ORDER BY
usr.username ASC,
ord.date DESC';
// Prepare and check sql statement (returns PDO statement).
$statement = $connection->prepare($sql);
if (!$statement) {
throw new Exception('The SQL statement can not be prepared!');
}
// Bind values to sql statement parameters.
$statement->bindValue(':takeawayid', $_SESSION['userid'], getInputParameterDataType($_SESSION['userid']));
$statement->bindValue(':paid', $paid, getInputParameterDataType($paid));
// Execute and check PDO statement.
if (!$statement->execute()) {
throw new Exception('The PDO statement can not be executed!');
}
// Fetch data.
$fetchedData = $statement->fetchAll(PDO::FETCH_ASSOC);
if ($fetchedData === FALSE) {
throw new Exception('Fetching data failed!');
}
return $fetchedData;
}
/**
* Group orders by users.
*
* @param array $orders [optional] Orders list.
* @return array Orders list grouped by users.
* @throws Exception
*/
function groupOrdersByUsers(array $orders = array()) {
$groupedOrders = array();
foreach ($orders as $order) {
$userId = $order['userId'];
$username = $order['username'];
$price = $order['price'];
// Check and add user name as key, if not already.
if (!array_key_exists($userId, $groupedOrders)) {
$groupedOrders[$userId] = array(
'username' => $username,
'orders' => array(),
'ordersPrice' => 0,
);
}
// Add order to grouped orders list.
$groupedOrders[$userId]['orders'][] = $order;
$groupedOrders[$userId]['ordersPrice'] += $price;
}
return $groupedOrders;
}
我是否需要一个额外的列，可能是 orderid，然后让所有订单具有相同的 id，然后按订单 id 显示每个订单？问题是在订单中的每个项目中添加相同的数字，然后下一个订单添加一个新数字并且没有 2相同的数字
编辑 现在我像这样展示购物车
$statement = $db->prepare("SELECT * FROM orders WHERE user_id = ? AND status = ? AND takeawayid = ? ");
// $statement->execute(array($username2,$md5password,$mod));
$statement->bindParam(1, $_SESSION['userid'], PDO::PARAM_STR, 50); // check your lengths
$statement->bindParam(2, $testing, PDO::PARAM_STR, 60);
$statement->bindParam(3, $id1, PDO::PARAM_INT);
$statement->execute();
这当然是完美的，因为旧订单将被接受或拒绝，因此等待状态的订单将是当前订单。我只是发现很难打印出旧订单并将它们组合在一起
【问题讨论】：