how to reference a foreign key when making a command-CodePudding

I am trying to make a custom command that I can call up within my Django project but am running into an issue referencing the foreign key it references.

How do I correctly reference the created instance in order to complete the command? Any help would be appreciated.

models.py

class Client(models.Model):
    name = models.CharField(max_length=50)

class Project(models.Model):
    schema_name = models.CharField(max_length=50)
    client = models.ForeignKey(Client, on_delete=models.CASCADE)

command

class Command(BaseCommand):
    help = (
        "commands, --create_public"
    )

    def add_arguments(self, parser):

        parser.add_argument(
            '--create_public',
            action='store_true',
            help='creates public tenant'
        )

    def handle(self, *args, **options):

        if options['create_public']:
            # create your public tenant
            client = Client(name='client1')
            client.save()
            tenant = Project(schema_name='public',
                           client='client1',
                           )
            tenant.save()

error

ValueError: Cannot assign "client": "Project.client" must be a "Client" instance.

CodePudding user response：

For future reference:

In your class Project you defined the constraint as being of type Client so the interpreter expects and instance of that class in the __init__created by the used framework. Thus raising ValueError when passing string client1 (although an instance's name).

Solution: pass the instance client when creating an instance of Project.