Php - cannot create new database record because primary key is null (autoincrement)-CodePudding

There are two tables: address and person. person_id (primary key, autoincrement) is a foreign key to address. I am trying to add an address record and person record from the same form but am getting this error.

Uncaught Error: Attempt to assign property "person_id" on null in C:\xampp\htdocs\app\controllers\Main.php:22 Stack trace: #0 [internal function]: app\controllers\Main->insert() #1 C:\xampp\htdocs\app\core\App.php(52): call_user_func_array(Array, Array) #2 C:\xampp\htdocs\index.php(4): app\core\App->__construct() #3 {main} thrown in C:\xampp\htdocs\app\controllers\Main.php on line 22

    public function insert(){
        if(isset($_POST['action'])){
            $person = new \app\models\Person();
            $person->first_name =($_POST['first_name']);
            $person->last_name =($_POST['last_name']);
            $person->notes =($_POST['notes']);
            $person->insert();

            $address->person_id = $person->person_id; //this is line 22
            $address->description =($_POST['description']);
            $address->street_address =($_POST['street_address']);
            $address->city =($_POST['city']);
            $address->province =($_POST['province']);
            $address->zip_code =($_POST['zip_code']);
            $address->country_code =($_POST['country_code']);
            $address->insert();         
            header('location:/Main/index');
        }else
            $this->view('Main/insert');
    }

Person insert method

    public function insert(){
        $SQL = 'INSERT INTO Person(person_id, first_name, last_name, `notes`) VALUES (:person_id,:first_name, :last_name, :notes)';
        $STMT = self::$_connection->prepare($SQL);
        $STMT->execute(['person_id'=>$this->person_id, 'first_name'=>$this->first_name, 'last_name'=>$this->last_name,'notes'=>$this->notes]);
    }

Address insert

    public function insert(){
        $SQL = 'INSERT INTO Address(person_id, description, street_address, city, province, zip_code, country_code) VALUES (:person_id, :description, :street_address, :city, :province, :zip_code, :country_code)';
        $STMT = self::$_connection->prepare($SQL);
        $STMT->execute(['person_id'=>$this->person_id, 'description'=>$this->description, 'street_address'=>$this->street_address, 'city'=>$this->city, 'province'=>$this->province, 'zip_code'=>$this->zip_code, 'country_code'=>$this->country_code]);
    }

CodePudding user response：

You have to use mysqli insert_id, after executing the insert command. insert_id will give you the primary key of the insert.

After

$person->insert();
$address->person_id = $person->insert_id;

CodePudding user response：

At no time do you use your Address model to insert the data into your address table.

Modify your code by adding this line :

$address = new \app\models\Address(); //ADD THIS LINE
$address->person_id = $person->person_id;
//...