抓取当前登录用户的用户ID并将其作为外键值添加到另一个表中返回null - MySQL - CodeIgniter

I'm trying to insert data based on current logged in user. I'm grabbing the user email address from the session data, then I query users table for the user_id that matches the unique email address but upon inserting it throws Error Number: 1452.I'm trying to insert data based on current logg