Can you create a new Model instance without saving it to the database

Question

I want to create a whole bunch of instances of a model object in Laravel, then pick the optimal instance and save it to the database. I know that I can create an instance with Model::create([]), but that saves to the database. If possible I'd like to create a bunch of models, then only "create" the one that is best.

Is this possible?

I am using Laravel 5.0

Answer 1

You create a new model simply by instantiating it:

$model = new Model;

You can then save it to the database at a later stage:

$model->save();

Answer 2

You can create instances with Model::make(). It works the same way as create but it doesn't save it.

Whether or not this is best practice is another matter entirely.

Answer 3

Yes, it is possible different ways: you can use the mass assignment without saving.

Please remember to set first the $fillable property in your model.

WAY #1: using the method fill

$model = new YourModel;

$model->fill([
    'field' => 'value',
    'another_field' => 'another_value'
]);

$model->save();

WAY #2: using the constructor

$model = new YourModel([
    'field' => 'value',
    'another_field' => 'another_value'
]);

$model->save();

Laravel documentation: https://laravel.com/docs/8.x/eloquent

Answer 4

there is also a method you can call it statically to get new instance:

 $modelInstance =   $modelName::newModelInstance();

it takes array $attributes = [] as a parameter