Insert into one table twice in same script? - php

Question

I've asked this question before, but have changed my code since. I'm having trouble with this script which inserts form data into a table. The first insert creates a booking which stores the customer's contact details. The second insert takes the booking ref created in the first and creates a 'JOB' for the customer. The final insert is supposed to create a second 'JOB', the customer's return journey.
The first two inserts are running fine,
but it ignored the final one, the second JOB insert.
I have checked the table structures, and the data been passed to the script everything is okay, so the problem must be in the script (shown below) any help is greatly appreciated.
Is it correct to use one script to insert into the same table twice?
<?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
));
}
?>

Answer 1

It is incorrect for sure, as inserting the same data twice violates one of most important database architecture laws - Database Normalization principle
However, there is no technical issues with it. There is some mistake which you have to catch using the error message from mysql. To have it, add this line after connecting to PDO.
$db->setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION);
Please note that catching the actual error is the only way to debug SQL queries. Just watching the code makes no sense nor help.
return must be a mysql keyword. write it as
`return`
By the way, I can't stand such enormously huge code.
If I were you, I'd make it in 10 lines, not 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);

Answer 2

It looks like booking_ref is the primary key in the jobs table, your trying to insert the same key twice which is why the final query fails.
You should have a seperate field that is the primary key on jobs which is just an auto-incrementing number, then create an index on booking_ref.

Answer 3

There's no law against it. What you need to do is check the return value for the last INSERT query. My best guess is there's a unique index on the jobs table that you're violating with the double-insert.
It's not obvious if you're using mySQLi or PDO here, but both's execute functions return false on failure, so you should catch that and then call the respective object's error functions to get what went wrong.