ID from another table on insertion using PDO

I am creating a quiz based system which part of it includes 2 tables:

Answer_bank table:

+-------+---------+----------+
| ab_id | ab_name | ab_qb_id |
+-------+---------+----------+

and a Question_bank table:

+-------+-------------+
| qb_id | qb_question | 
+-------+-------------+

The aim is to allow someone to create a question and an answer, the answer will be stored within the answer bank table with the ab_qb_id equal to the qb_id. I don't want this in the same table as I will be making this more complex.

I try to use the following PDO/SQL to insert into both the tables.

//questions
$qb_id = $_POST['qb_id'];
$qb_question = $_POST['qb_question'];
$sql = "INSERT INTO questions_bank (`qb_id`, `qb_question`)
        VALUES (:qb_id, :qtn)";
$stmt = $db->prepare($sql);
$stmt->bindValue(":qb_id", $qb_id);
$stmt->bindValue(":qtn", $qb_question);
$stmt->execute();

//answers
$ab_name = $_POST['ab_name'];
$sql = "INSERT INTO answers_bank (`ab_name`, `ab_qb_id`) VALUES (:ab_name, :qb_id)";
$stmt = $db->prepare($sql);
$stmt->bindValue(':ab_name', $ab_name);
$stmt->bindValue(':qb_id', $qb_id);
$stmt->execute();

However the problem I have is the ab_qb_id in the answer_bank table always inserts 0 and not the same id as qb_id. Is this the incorrect way to do this? What's the best way for the answer table to include the qb_id ?... So that then the answer is related to a specific question. Thank you

I would personally get the ID of the row that was just inserted in the questions_bank table and then use that ID as the value you insert into the answers_bank.ab_qb_id column. Because you're using PDO, you can use this: $db->lastInsertId(). For example:

$qb_id = $_POST['qb_id'];
$qb_question = $_POST['qb_question'];
$sql = "INSERT INTO questions_bank (`qb_id`, `qb_question`)
    VALUES (:qb_id, :qtn)";
$stmt = $db->prepare($sql);
$stmt->bindValue(":qb_id", $qb_id);
$stmt->bindValue(":qtn", $qb_question);
$stmt->execute();
$inserted_id = $db->lastInsertId();

//answers
$ab_name = $_POST['ab_name'];
$sql = "INSERT INTO answers_bank (`ab_name`, `ab_qb_id`) VALUES (:ab_name, :qb_id)";
$stmt = $db->prepare($sql);
$stmt->bindValue(':ab_name', $ab_name);
$stmt->bindValue(':qb_id', $inserted_id); //Use the previously inserted ID
$stmt->execute();

Using this method will ensure that the questions_bank.qb_id and answers_bank.ab_qb_id are the same. To make sure that any interruptions (power surge, disk failure, etc.) don't have a chance to affect this, you can wrap these in a transaction. Then you can be sure that the values will always match.

EDIT I forgot to add the try{}catch{} statement to the transaction:

For example:

try{
    $db->beginTransaction();
    //Your current queries
    $db->commit();
}catch(Exception $e){
    $db->rollback();
    die($e->getMessage());
}