How not to save record if id already exists in laravel?

Question

MySQL

foreach ($items as $item) {
            $carObj = new Car();

            $carObj->id = $item['id'];
            $carObj->make = $item['make'];
            $carObj->model = $item['model'];
            $carObj->price = $item['price'];
            $carObj->href = $item['href'];
            $carObj->year = $item['year'];
            $carObj->created = $item['created'];
            $carObj->save();
        }

How not to write a record if such id already exists?

Answer 1

How not to write a record if such id already exists?

Make a try catch

try {
            $carObj = new Car();
            $carObj->id = $item['id'];
            $carObj->make = $item['make'];
            $carObj->model = $item['model'];
            $carObj->price = $item['price'];
            $carObj->href = $item['href'];
            $carObj->year = $item['year'];
            $carObj->created = $item['created'];
            $carObj->save();
} catch (Illuminate\Database\QueryException $e) {
    /* nothing */
}

This way we catch the exception and just skip it. An error in this case will not stop the script.

Answer 2

At the moment, if id exists, then an error is displayed

Answer 3

Do you need to save the id? You don't have autoincrement in the database?

Answer 4

if ($car = Car::firstOrNew(['id' => $item['id']) and !$car->exists) {
    $car->make = $item['make'];
    ...
    $car->save();
}