Php 如何在同一个表中插入两次？

我正在创建一个预订系统（机场出租车公司），该系统最初创建一个预订，存储客户联系信息和账单详细信息等，然后使用ID将旅程信息发布到另一个表中，如果包含了回程详细信息，则会在回程的同一个工作表中创建另一个记录。我使用IF语句检查是否已提交退货详细信息。然而，它只插入第一次旅程，而不是返回。请参阅下面的代码

有什么想法吗

我也很欣赏，我很可能会以一种奇怪的方式来实现我的结果，所以任何批评和指点都是受欢迎的

谢谢

<?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'];
$pickup_destination      =  $_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;
$account_number          =  9999;
$authorised              =  N;


$booking_date            = 0;
$null_date               = '0000-00-00'; 


include('../assets/db_connection.php');

mysql_select_db("airporthopper");

mysql_query("INSERT INTO bookings(customer_name, billing_address, account_number, contact_tel, contact_mob, contact_email,
            party_pax, party_cases, booking_notes, price, booking_agent, booking_date, booking_status, authorised)
VALUES('$customer_title $customer_first_name $customer_last_name', '$billing_address', $account_number, '$customer_tel', '$customer_mobile', '$customer_email', '$passengers', '$cases', '$booking_notes', '$price', 'Danny Green', '$booking_date',
    '$booking_status', '$authorised'   )");



$booking_ref = mysql_insert_id();

mysql_query("INSERT INTO jobs(booking_ref, pickup_date, pickup_time, pickup_address, destination_address, scheduled)VALUES('$booking_ref', '$pickup_date', '$pickup_time', '$pickup_address', '$pickup_destination', 'N')");

if ($return_date != $null_date) {
mysql_query("INSERT INTO jobs(booking_ref, pickup_date, pickup_time, pickup_address, destination_address, return, scheduled)VALUES('$booking_ref', '$return_date', '$return_time', '$return_pickup', '$return_destination', 'Y' , 'N')");
 }




?> 

使用以下命令代替“如果（$return\u date！=$null\u date）{”

另外，我希望“jobs”表中的“booking\u ref”列没有与之关联的唯一索引。
。它们不再被维护。请参阅？改为了解，并使用或-将帮助您决定哪一个。如果您选择PDO，。@johncode:您是否使用一些不引人注目的浏览器扩展或等效功能将其快速粘贴为snippet？只要我知道怎么做，我就想加入。@Jon.我从Evernote获得文本并使用Evernote进行剪切和粘贴。不要像别人说的那样使用mysql函数，但更重要的是不要在查询中使用未替换的POST变量。这是一个主要的安全威胁。试着检查你是否进入if（$return\u date！=$null\u date）{}或者不是。
if (strtotime($return_date) > 0) {