Laravel 5: Chicken and egg when creating a one-to-one relationship - laravel

Question

Let's assume I have two Model elements:
Person
Computer
In my application a Person May have one computer (a person has 0..1 computers).
So
In my User Model, I have:
public function computer()
{
return $this->hasOne('App\Model\Computer');
}
In my Computer Model I have:
public function owner_user() {
return $this->belongsTo('App\Model\User');
}
Sometimes in my application I need to create a computer. computers cannot be without owners, so when I get the information about the computer to create, I also get information about the owner.
What is the most elegant way to create BOTH new objects (computer and person), associate them with the reference IDs without doing multiple saves?
My problem is: the IDs are only being generated when the eloquent model is saved. because of that, I have a chicken and egg problem where I can't save a computer, because I don't know the ID of the owner person, and I can't save a person because I don't know the ID of the owned computer.
I could only do this by doing multiple saves and having a stub ID temporarily saved until the other objects is saved, and then I update the ID of the first one saved.
There must be a more elegant way to save two or more items at the same time, with reference IDs in place, with a single call - and without having to worry about database ID integrity in a high volume web server.
It seems the associate function is for existing records. Is there no other way for my situation than having to save person + save computer + associate?
Here's the schema of my people (users) and computers.
users
+----------------+---------------------+------+-----+---------+----------------+
| Field | Type | Null | Key | Default | Extra |
+----------------+---------------------+------+-----+---------+----------------+
| id | bigint(20) unsigned | NO | PRI | NULL | auto_increment |
| username | varchar(45) | NO | UNI | NULL | |
| email | varchar(45) | NO | | NULL | |
| password | text | NO | | NULL | |
| remember_token | varchar(100) | NO | | NULL | |
| created_at | timestamp | YES | MUL | NULL | |
| updated_at | timestamp | YES | MUL | NULL | |
+----------------+---------------------+------+-----+---------+----------------+
computers
+----------------------------+---------------------+------+-----+---------+----------------+
| Field | Type | Null | Key | Default | Extra |
+----------------------------+---------------------+------+-----+---------+----------------+
| id | bigint(20) unsigned | NO | PRI | NULL | auto_increment |
| user_id | bigint(20) | NO | UNI | NULL | |
| name | varchar(45) | NO | MUL | NULL | |
| created_at | timestamp | YES | MUL | NULL | |
| updated_at | timestamp | YES | MUL | NULL | |
+----------------------------+---------------------+------+-----+---------+----------------+

Answer 1

For the least amount of database work, you want to create the user first and then create the computer. This way, you can have one save for the user, and then one save for the computer, which is only two writes to the database.
If you create the computer first, you would have to create the computer, create the user, and then update the computer with the user id, which would be three writes to the database instead of two.
However, since you're creating two new records (user and computer), the fact is that you're going to have at least two writes to the database, no matter what you do.
This is what you're looking at:
// create the user object in the database (first write)
$user = \App\Model\User::create([
'name' => $request->input('user_name'),
'password' => $request->input('user_password')
]);
// the save on the relationship will update the foreign key on the
// given model and save the whole model to the database (second write).
$user->computer()->save(new \App\Model\Computer([
'name' => $request->input('computer_name')
]));
It sounds like another concern you may have is the referential integrity. What happens if the user saves fine, but there is a problem saving the computer? Do you want to keep the user, or should it seem like the user save never happened?
If you have a multiple database writes that should be treated as an all-or-nothing situation, you'll need a database transaction. If your work is done inside a database transaction, you can have all the database writes automatically rolled back if one of them fails.
DB::transaction(function() {
$user = \App\Model\User::create([
'name' => $request->input('user_name'),
'password' => $request->input('user_password')
]);
$user->computer()->save(new \App\Model\Computer([
'name' => $request->input('computer_name')
]));
});
With this code, if the save of the computer throws an exception, then the user that was created will automatically be undone. Laravel's documentation on transactions is here.
One thing to note regarding transactions is that you need to make sure the database you're using supports them. For example, for MySQL, only the InnoDB and BDB storage engines support transactions. All other storage engines (e.g. MyISAM) do not. The code will not throw any errors, but the database writes that complete will not be rolled back.

Answer 2

This can be achieved easily with the push() method.
Sometimes you may wish to save not only a model, but also all of its relationships. To do so, you may use the push method:
Saving A Model And Relationships
$user->push();
Example code:
$user = new Person();
$user->email = Request::get('email');
$user->password = Request::get('password');
$user->computer = new Computer();
$user->computer->name = Request::get('computer_name');
$user->push();
You can read more about Eloqeunt relations in the documentation.

Answer 3

You are not suppose to have foreign ids in both tables.
You should have user_id only on computers table.
So you will create a user and then a computer.