python - Django error. Cannot assign must be an instance

Question

I get the following error when I try to run an insert into one of my tables.

Cannot assign "1": "Team.department_id" must be a "Department" instance

Admittedly I'm slightly unsure if I'm using the foreign key concept correctly. The insert I'm trying to run and a snippet from my models.py are given below.

What I'm trying to do is that when someone wants to create a new team. They have to attach it to a department. Therefore the department ID should be in both sets of tables.

new_team = Team(
    nickname = team_name,
    employee_id = employee_id,
    department_id = int(Department.objects.get(password = password, department_name = department_name).department_id)
)

models.py

class Department(models.Model):  
    department_id = models.AutoField(auto_created=True, primary_key=True, default=1)  
    department_name = models.CharField(max_length=60)
    head_id = models.CharField(max_length=30)
    password = models.CharField(max_length=128)


class Team(models.Model):  
    team_id = models.AutoField(primary_key=True)
    department_id = models.ForeignKey('Department', related_name = 'Department_id')
    employee_id = models.CharField(max_length=30)
    nickname = models.CharField(max_length=60)
    team_image = models.ImageField(upload_to=get_image_path, blank=True, null=True)

Answer 1

You don't need to pass the department id, the instance itself is enough. The following should work just fine:

new_team = Team(
    nickname = team_name,
    employee_id = employee_id,
    department_id = Department.objects.get(password = password, department_name = department_name)
)

Just a note, don't ever name your foreign fields something_id. That something is enough. Django is meant to make things easy from the user's perspective and the _id suffix means you're thinking of the database layer. In fact, if you named your column department, django will automatically create department_id column in the database for you. The way things are, you're making django create department_id_id which is rather silly.