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两个错误-在自定义管理面板上将项目标记为随附按钮

Two Errors - Mark an item as shipped with button on custom admin panel

本文关键字：项目 目标 记为 按钮 管理 两个 错误 自定义 | 更新日期: 2023-09-27

我用自己的数据库制作了一个管理面板。

在连接到另一个（自己的）数据库的preorder.php页面上，在用户填写表格并对订单进行条纹处理后，我会检查是否像这样支付了费用；

if ($charge->paid == true) {
    $amountReadable = $amount / 100; // to add in decimal points
    echo '<div class="alert alert-success">Your card was successfully billed for $'.$amountReadable.'</div>';
    $status = "paid";

然后我连接到主数据库和管理数据库（它们都有"订单"表），并插入以下查询：

 $connect = mysql_connect("localhost",DB_USER,DB_PASS);
 if (!$connect){
 die('Could not connect: ' . mysql_error());
 }
 mysql_select_db(DB_NAME, $connect);
 $query = "INSERT INTO 'DB_NAME'.`orders` (`email`, `name`, `qty`, `product`, `amount`, `stripe_customer_id`, `stripe_charge_id`, `address1`, `address2`, `city`, `state`, `zip`, `timestamp`, `status`) VALUES ('$email','$cardName', '$qty', '$product', '$amountReadable', '$customer->id', '$charge->id', '$cardAddress1', '$cardAddress2', '$cardCity', '$cardState', '$cardZipcode', CURRENT_TIMESTAMP`, $status);";
 mysql_query($query);
 if (mysql_errno()) {
 $error = "MySQL error ".mysql_errno().": ".mysql_error()."'n<br>When executing:<br>'n$query'n<br>";
 exit;   
 }
 mysql_close($connect);
 //insert into db for admin
 $connect = mysql_connect("--------","------","-------");
 if (!$connect){
 die('Could not connect: ' . mysql_error());
 }
 mysql_select_db(database2, $connect);
 $query = "INSERT INTO 'database2'.`orders` (`email`, `name`, `qty`, `product`, `amount`, `stripe_customer_id`, `stripe_charge_id`, `address1`, `address2`, `city`, `state`, `zip`, `timestamp`, `status`) VALUES ('$email','$cardName', '$qty', '$product', '$amountReadable', '$customer->id', '$charge->id', '$cardAddress1', '$cardAddress2', '$cardCity', '$cardState', '$cardZipcode', CURRENT_TIMESTAMP, $status);";

然后在有自己数据库的管理面板（上面的数据库2）上，（orders.php）我将这些数据拉到一个引导表中，如下所示：

编辑1#：我在管理面板中将orders.php更改为类似的内容：

$result = mysqli_query($con,"SELECT * FROM orders");
echo "<table border='1' data-toggle='table'>
<tr>
<th>id</th>
<th>email</th>
<th>name</th>
<th>qty</th>
<th>product</th>
<th>amount</th>
<th>address1</th>
<th>address2</th>
<th>city</th>
<th>state</th>
<th>zip</th>
<th>status</th>
</tr>";
 while($row = mysqli_fetch_array($result))
 {
 echo "<tr>";
  echo "<td>" . $row['id'] . "</td>";
  echo "<td>" . $row['email'] . "</td>";
  echo "<td>" . $row['name'] . "</td>";
  echo "<td>" . $row['qty'] . "</td>";
  echo "<td>" . $row['product'] . "</td>";
  echo "<td>" . $row['amount'] . "</td>";
  echo "<td>" . $row['address1'] . "</td>";
  echo "<td>" . $row['address2'] . "</td>";
  echo "<td>" . $row['city'] . "</td>";
  echo "<td>" . $row['state'] . "</td>";
  echo "<td>" . $row['zip'] . "</td>";
  echo "<td>" . $row['status']. "</td>";
  echo "<td><form action='markshipped.php' method='POST'><input type='hidden' name='status' value='".$row["id"]."'/><input type='submit' name='submit-btn' value='Mark Item Shipped' /><form></td>";
  echo "</tr>";
 }
echo "</table>";

那么我的markshipped.php是：

<?php
$con=mysqli_connect("----","----","----","----");
// Check connection
if (mysqli_connect_errno()) {
  echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
mysqli_query($con,"UPDATE orders SET status='shipped' WHERE id=$id");
mysqli_close($con);
?>

我仍然对php/mysql不太满意，不幸的是，在有人购买任何东西之前，我无法对其进行测试，因为在条纹测试模式下，没有一张卡无法通过表单…

所以我的问题主要是：

我收到两个错误，说：

1.未定义的变量：第9行上/admin/orders/markshipped.php中的id

-这是哪条线路：mysqli_query($con,"UPDATE orders SET status='shipped' WHERE id=$id");

2.未定义的索引：第88行/admin/orders/orders.php中的状态

-即：echo "<td>" . $row['status']. "</td>";

很明显，我把$status = "paid";设置错了和/或放错了地方。

如何正确设置这些选项，以便在开始时将项目标记为"已付款"，然后在单击每行上的"标记项目已发货"按钮后将其更改为"已发货"？你能在我的代码中发现其他错误吗

非常感谢您抽出时间！

您需要使用获取$_POST[状态]

$id = $_POST['status'];

以便运行您的查询。我会将你的隐藏输入字段的名称更改为"id"，并获取$_POST["id"]，因为你想提交id。

您在这一行中有一个`（第一个插入查询）：

CURRENT_TIMESTAMP`<<<<here , $status

并且需要删除这个/更好地在mysql 中将列设置为on_update_current_timestamp

$status需要在"中，我假设$id是一个整数，所以可以不使用"。

如果您混合了mysql_和mysqli_，我强烈建议您在mysqli_中重写整个代码，特别是对于插入语句，您确实应该使用prepared语句和mysqli_。mysql贬值且不安全。您对sql注入持开放态度

为什么将id设置为NULL？这应该是一个自动递增字段，您不应该在查询中将其设置为null。

如果你把对象数据放在数据库中，我会使用'{$customer->id}'或设置

$customer_id = $customer->id;

在查询之前，更容易检查您的查询

您不需要指定

INSERT INTO **'database2'**.订单

database2，数据库已被选中-再次：请使用mysqli_和prepared语句！

示例：

$customer_id = $customer->id;
$charge_id = $charge->id;
$stmt = $sql->prepare("INSERT INTO `orders` (`email`, `name`, `qty`, `product`, `amount`, `stripe_customer_id`, `stripe_charge_id`, `address1`, `address2`, `city`, `state`, `zip`, `status`) VALUES (?,?,?,?,?,?,?,?,?,?,?,?)");
$stmt->bind_param('ssissssssssss', $email,$cardName, $qty, $product, $amountReadable, $customer_id, $charge_id, $cardAddress1, $cardAddress2, $cardCity, $cardState, $cardZipcode, $status);
$stmt->execute();

如果您这样做，则需要为mysqli_而不是为mysql_设置连接

区块1:

<?php

if ($charge->paid == true) {
    $amountReadable = $amount / 100; // to add in decimal points
    echo '<div class="alert alert-success">Your card was successfully billed for $' . $amountReadable . '</div>';
    $status = "paid";
}

区块2:

<?php
$connect = new mysqli("localhost",DB_USER,DB_PASS,DB_NAME);
$customer_id = $customer->id;
$charge_id = $charge->id;
$stmt = $connect->prepare("INSERT INTO orders (`email`, `name`, `qty`, `product`, `amount`, `stripe_customer_id`, `stripe_charge_id`, `address1`, `address2`, `city`, `state`, `zip`, `status`) VALUES (?,?,?,?,?,?,?,?,?,?,?,?,?)");
$stmt->bind_param('ssissssssssss', $email,$cardName, $qty, $product, $amountReadable, $customer->id, $charge->id, $cardAddress1, $cardAddress2, $cardCity, $cardState, $cardZipcode, $status);
$stmt->execute();


$connect->close();
//insert into db for admin
$connect = new mysqli("--------","------","-------","--------");
$customer_id = $customer->id;
$charge_id = $charge->id;
$stmt = $connect->prepare("INSERT INTO `orders` (`email`, `name`, `qty`, `product`, `amount`, `stripe_customer_id`, `stripe_charge_id`, `address1`, `address2`, `city`, `state`, `zip`, `status`) VALUES (?,?,?,?,?,?,?,?,?,?,?,?)");
$stmt->bind_param('ssissssssssss', $email,$cardName, $qty, $product, $amountReadable, $customer_id, $charge_id, $cardAddress1, $cardAddress2, $cardCity, $cardState, $cardZipcode, $status);
$stmt->execute();

块3:

$con = new mysqli("--------","------","-------","--------");
$query = "SELECT * FROM orders";
if ($result = $con->query($query))
{
echo "<table border='1' data-toggle='table'>
<tr>
<th>id</th>
<th>email</th>
<th>name</th>
<th>qty</th>
<th>product</th>
<th>amount</th>
<th>address1</th>
<th>address2</th>
<th>city</th>
<th>state</th>
<th>zip</th>
<th>status</th>
</tr>";
while($row = $result->fetch_assoc()) {
    echo "<tr>";
    echo "<td>" . $row['id'] . "</td>";
    echo "<td>" . $row['email'] . "</td>";
    echo "<td>" . $row['name'] . "</td>";
    echo "<td>" . $row['qty'] . "</td>";
    echo "<td>" . $row['product'] . "</td>";
    echo "<td>" . $row['amount'] . "</td>";
    echo "<td>" . $row['address1'] . "</td>";
    echo "<td>" . $row['address2'] . "</td>";
    echo "<td>" . $row['city'] . "</td>";
    echo "<td>" . $row['state'] . "</td>";
    echo "<td>" . $row['zip'] . "</td>";
    echo "<td>" . $row['status'] . "</td>";
    echo "<td><form action='markshipped.php' method='POST'><input type='hidden' name='id' value='" . $row["id"] . "'/><input type='submit' name='submit-btn' value='Mark Item Shipped' /></form></td>";
    echo "</tr>";
}}
echo "</table>";

区块4:

$con= new mysqli("----","----","----","----");
// Check connection
$id = $_POST['id'];
    $stmt = $con->prepare("UPDATE orders SET status = 'shipped' WHERE id = ?");
    $stmt->bind_param('i', $id);
    $stmt->execute();
$con->close();

对于您的第一个问题：

mysqli_query（$con，"UPDATE orders SET status='shipped'WHERE id=$id"）；

WHERE id=$id应为：WHERE id='$id'（在$id两侧添加单引号）。

第二个问题：

…'$cardZipcode'，CURRENT_TIMESTAMP'，$status）；"；

同样的事情：CURRENT_TIMESTAMP, $status);应该是：CURRENT_TIMESTAMP, '$status');（你需要在两个地方纠正这个）。

我也经常犯这些错误。

我希望这能有所帮助！

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