Django模型中的外键和主键关系

Question

These models are in two different applications in django. I had created the primary key and foreign key relation between the two tables, but the output is not desired output.

class User(models.Model):
    user_name = models.CharField( max_length=255, default=None)
    email = models.CharField( max_length=255, default=None)
    mobile_number = models.IntegerField( default=0)
    created_on = models.DateTimeField( auto_now=True)

class Todo(models.Model):
    title = models.CharField( max_length=255, default=None)
    description = models.CharField( max_length=255, default=None)
    created_on = models.DateTimeField(auto_now=True)
    user = models.ForeignKey('Account.User',on_delete=models.CASCADE)

Suppose I created a entry in the User table and the id assigned is 1 , but I can also create a entry in the Todo table with user_id = 2 (which is not a valid value according to the rule ), and the entry is created. Can any one suggest why?

In Python shell:

from Account.models import User
my_user = ( user_name = "Abhis", email = "aa@aa.com", mobile_number = "9999999999", created_on = datetime.datetime.today())
my_user.save()    

# it will create a row with id 1.
from ToDo.models import Todo
my_todo = ( title = "title", description = "description", created_on = datetime.datetime.today(), user_id = 4)
my_todo.save()
# the row with this user_id = 4 is created, which must be illegal 

Answer 1

The django documentation states that this will not happen and it will throw DoesNotExist exception. Kindly check the link below

Django raises DoesNotExist when consulting an empty ForeignKey field or key that does not exist

But if you are connecting to a database that exists, and it has values which are not proper foreign keys that is also mentioned in the below link. It clearly states that the model is wrong and should be changed in such a scenario.Link below

documentation: invalid foreign key and select_related

When i tried to assign a random value to a foreign key it threw ValueError. Can you check if your DB does not have an entry with user_id=4 .