我以前问过这个问题，但后来更改了代码。我在使用这个将表单数据插入表的脚本时遇到了问题。第一个插入创建一个预订，其中存储客户的联系方式。第二个插入采用在第一个插入中创建的预订引用，并为客户创建一个"JOB"。最后一个插入应该创建第二个"JOB"，即客户的回程。
前两个插件运行良好，但它忽略了最后一个，即第二个JOB插入。
我已经检查了表结构，数据已经传递到脚本中，一切都很好，所以问题一定在脚本中（如下所示），非常感谢任何帮助。
使用一个脚本在同一个表中插入两次是否正确？
<?php
$customer_title = $_POST['customer_title'];
$customer_first_name = $_POST['customer_first_name'];
$customer_last_name = $_POST['customer_last_name'];
$billing_address = $_POST['billing_address'];
$customer_tel = $_POST['customer_tel'];
$customer_mobile = $_POST['customer_mobile'];
$customer_email = $_POST['customer_email'];
$passengers = $_POST['passengers'];
$cases = $_POST['cases'];
$return_flight_number = $_POST['return_flight_number'];
$price = $_POST['price'];
$pickup_date = $_POST['pickup_date'];
$pickup_time = $_POST['pickup_time'];
$pickup_address = $_POST['pickup_address'];
$destination_address = $_POST['pickup_destination'];
$return_date = $_POST['return_date'];
$return_time = $_POST['return_time'];
$return_pickup = $_POST['return_pickup'];
$return_destination = $_POST['return_destination'];
$booking_notes = $_POST['booking_notes'];
$booking_status = "Confirmed";
$authorised = "N";
$booking_agent = "ROOT_TEST";
$booking_date = date("Y/m/d");
if (isset($_POST['customer_title'])) {
include('../assets/db_connection.php');
$create_booking = $db->prepare("INSERT INTO bookings(customer_name, billing_address, contact_tel, contact_mob, contact_email, party_pax, party_cases, booking_notes, price, booking_agent, booking_date, booking_status, authorised)
VALUES(:customer_name, :billing_address, :contact_tel, :contact_mob, :contact_email, :party_pax, :party_cases, :booking_notes, :price, :booking_agent, :booking_date, :booking_status, :authorised );");
$create_booking->execute(array(
":customer_name" => $customer_title . ' ' . $customer_first_name . ' ' . $customer_last_name,
":billing_address" => $billing_address,
":contact_tel" => $customer_tel,
":contact_mob" => $customer_mobile,
":contact_email" => $customer_email,
":party_pax" => $passengers,
":party_cases" => $cases,
":booking_notes" => $booking_notes,
":price" => $price,
":booking_agent" => $booking_agent,
":booking_date" => $booking_date,
":booking_status" => $booking_status,
":authorised" => $authorised
));
$booking_ref = $db->lastInsertId('booking_ref'); // Takes Booking Ref generated in $create_booking
$scheduled = "N";
$create_job = $db->prepare("INSERT INTO jobs(booking_ref, pickup_date, pickup_time, pickup_address, destination_address, scheduled)
VALUES(:booking_ref, :pickup_date, :pickup_time, :pickup_address, :destination_address, :scheduled)");
$create_job->execute(array(
":booking_ref" => $booking_ref,
":pickup_date" => $pickup_date,
":pickup_time" => $pickup_time,
":pickup_address" => $pickup_address,
":destination_address" => $destination_address,
":scheduled" => $scheduled
));
$return = "Y";
$create_return = $db->prepare("INSERT INTO jobs(booking_ref, pickup_date, pickup_time, pickup_address, destination_address, scheduled, return)
VALUES(:booking_ref, :pickup_date, :pickup_time, :pickup_address, :destination_address, :scheduled, :return)");
$create_return->execute(array(
":booking_ref" => $booking_ref,
":pickup_date" => $return_date,
":pickup_time" => $return_time,
":pickup_address" => $return_pickup,
":destination_address" => $return_destination,
":scheduled" => $scheduled,
":return" => $return
));
}
?>
这肯定是不正确的，因为两次插入相同的数据违反了最重要的数据库体系结构定律之一——数据库规范化原则
但是，它没有技术问题。您必须使用mysql中的错误消息来捕捉一些错误。要获得它，请在连接到PDO后添加此行。
$db->setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION);
请注意，捕捉实际错误是调试SQL查询的唯一方法。仅仅看代码没有任何意义，也没有任何帮助。
return必须是mysql关键字。将其写入
`return`
顺便说一句，我无法忍受如此巨大的代码。
如果我是你，我会用10行，而不是50:
$allowed = array('customer_name', 'billing_address', 'contact_tel', 'contact_mob',
'contact_email', 'party_pax', 'party_cases', 'booking_notes', 'price');
$insert = $db->filterArray($_POST,$allowed);
$insert['booking_status'] = "Confirmed";
$insert['authorised'] = "N";
$insert['booking_agent'] = "ROOT_TEST";
$insert['booking_date'] = date("Y-m-d");
$db->query("INSERT INTO bookings SET ?u", $insert);
看起来booking_ref是jobs表中的主键，您尝试插入同一个键两次，这就是最终查询失败的原因。
您应该有一个单独的字段，它是jobs上的主键，它只是一个自动递增的数字，然后在booking_ref上创建一个索引。
没有反对它的法律。您需要做的是检查最后一个INSERT查询的返回值。我的最佳猜测是jobs表上有一个唯一的索引，您使用双插入违反了该索引。
这里使用的是mySQLi还是PDO并不明显，但两者的执行函数都会在失败时返回false，因此您应该捕捉到这一点，然后调用相应对象的错误函数来了解问题所在。