Php 在同一脚本中插入一个表两次？

我以前问过这个问题，但后来更改了代码。这个脚本将表单数据插入到表中，我遇到了问题。第一次插入创建一个预订，其中存储客户的联系方式。第二个insert获取在第一个insert中创建的booking ref，并为客户创建一个“JOB”。最后的插入应该创建第二个“工作”，即客户的回程

前两个插件运行良好， 但它忽略了最后一个，第二个作业插入

我已经检查了表结构，数据已经传递到脚本，一切正常，所以问题一定在脚本中（如下所示）。非常感谢您的帮助

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
  ));




}


?>

---

看起来

booking\u ref

是

jobs

表中的主键，您试图插入相同的键两次，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

作业

上的主键，它只是一个自动递增的数字，然后在

预订_ref

上创建一个索引。您需要做的是检查最后一次插入查询的返回值。我最好的猜测是，在

jobs

表中有一个唯一的索引，您使用双插入违反了该索引

---

这里是否使用mySQLi或PDO并不明显，但两个执行函数在失败时都返回

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以获取错误信息。

这肯定是错误的，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规则之一-原则

但是，它没有任何技术问题。使用mysql的错误消息，您必须捕捉到一些错误。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，捕获实际错误是调试SQL查询的唯一方法。仅仅看代码没有意义，也没有帮助

return必须是mysql关键字。写成

`return`

顺便说一句，我受不了这么庞大的代码

如果我是你，我会写10行，而不是50行：

$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);

---

在执行第三个查询之前，您是否可以对post值进行vardump，并查看它们是否包含正确的数据？是否将

PDO:：ATTR\u ERRMODE

设置为

PDO:：ERRMODE\u WARNING

（启用了

E\u WARNING

）或

PDO:：ERRMODE\u EXCEPTION

？作业上有一个主键，即job\u ref，我已经检查了phpmyadmin，唯一的主键是job\u ref，booking\u ref没有设置为unique。谢谢，你是对的，它是返回字段名，没有``任何一方都不喜欢。我还将考虑按照您的建议减少代码。再次感谢：）