INSERT INTO not inserting for me - php

Question

I can't see anything wrong with this, but it just wont work.
Quick rundown: Form filled in on previous page, $_POST['']'s added to $_SESSION['']'s, $vars set from the $_SESSION['']'s, $vars used to mysqli_query.
Im sure the problem is staring me in the face, but I just cant see it.
Heres the code: (UPDATED)
if(isset($_POST['list_make']) && $_POST['list_make'] != '') { $list_make = $_POST['list_make']; $_SESSION['list_make'] = $list_make; }
if(isset($_SESSION['list_make']) && $_SESSION['list_make'] != '') { $list_make = $_SESSION['list_make']; } else { $list_make = ''; }
$add_car_query = "INSERT INTO car_details
(car_user_id, car_user_number, car_date_added, car_make, car_model, car_date_registered, car_odometer,
car_engine_size, car_color, car_body_type, car_owners, car_nct_date, car_rax_date)
VALUES
('$user_id_new', '$new_user_number', '$today', '$list_make', '$list_model', '$list_year', '$list_kilometers',
'$list_engine_size', '$list_color', '$list_body_type', '$list_previous_owners', '$list_nct', '$list_tax')
";
if(mysqli_query($con, $add_car_query)) { $added = 'added'; } else { $added = 'Not happening'; }

Answer 1

Try checking if there is a MySQLi error in the else (which will be triggered if the query fails):
$add_car_query = "INSERT INTO car_details
(car_user_id, car_user_number, car_date_added, car_make, car_model, car_date_registered, car_odometer,
car_engine_size, car_color, car_body_type, car_owners, car_nct_date, car_rax_date)
VALUES
('$user_id_new', '$new_user_number', '$today', '$list_make', '$list_model', '$list_year', '$list_kilometers',
'$list_engine_size', '$list_color', '$list_body_type', '$list_previous_owners', '$list_nct', '$list_tax')
";
if(mysqli_query($con, $add_car_query)) {
$added = 'added';
}
else {
// check for the error here
if(mysqli_error()) {
echo mysqli_error();
}
$added = 'Not happening';
}
Prepared Statements
As someone in the comments said, you should use prepared statements for better security. This way you don't have to worry about escaping input (it looks like you aren't escaping the values right now which is very dangerous).
It can get a bit confusing with the question mark placeholders though (MySQLi doesn't support named parameters unfortunately). Prepared statements are nicer with OOP MySQLi, but you're using procedural so here's how you do a prepared statement for the query in your question:
$add_car_query = "INSERT INTO car_details
(car_user_id,
car_user_number,
car_date_added,
car_make,
car_model,
car_date_registered,
car_odometer,
car_engine_size,
car_color,
car_body_type,
car_owners,
car_nct_date,
car_rax_date)
VALUES (?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?)";
$stmt = mysqli_stmt_prepare($con, $add_car_query);
// if any of the values are an int, change the
// corresponding 's' (which means string) to 'i' (integer).
// more info: http://us.php.net/manual/en/mysqli-stmt.bind-param.php
mysqli_stmt_bind_param($stmt, 'sssssssssssss',
$user_id_new,
$new_user_number,
$today,
$list_make,
$list_model,
$list_year,
$list_kilometers,
$list_engine_size,
$list_color,
$list_body_type,
$list_previous_owners,
$list_nct,
$list_tax);
$stmt_exe = mysqli_stmt_execute($stmt);
if($stmt_exe) {
$added = 'added';
}
else {
if(mysqli_error()) {
echo mysqli_error();
}
$added = 'Not happening';
}