Django error message “Add a related_name argument to the definition”

Question

D:\\zjm_code\\basic_project>python manage.py syncdb
Error: One or more models did not validate:
topics.topic: Accessor for field \'content_type\' clashes with rel         


        

Answer 1

Example:

class Article(models.Model):
    author = models.ForeignKey('accounts.User')
    editor = models.ForeignKey('accounts.User')
    

This will cause the error, because Django tries to automatically create a backwards relation for instances of accounts.User for each foreign key relation to user like user.article_set. This default method is ambiguous. Would user.article_set.all() refer to the user's articles related by the author field, or by the editor field?

Solution:

class Article(models.Model):
    author = models.ForeignKey('accounts.User', related_name='author_article_set')
    editor = models.ForeignKey('accounts.User', related_name='editor_article_set')

Now, for an instance of user user, there are two different manager methods:

1. user.author_article_set — user.author_article_set.all() will return a Queryset of all Article objects that have author == user

2. user.editor_article_set — user.editor_article_set.all() will return a Queryset of all Article objects that have editor == user

---

Note: This is an old example — on_delete is now another required argument to models.ForeignKey.

Answer 2

I had a similar problem when I was trying to code a solution for a table that would pull names of football teams from the same table. My table looked like this:

hometeamID = models.ForeignKey(Team, null=False, on_delete=models.CASCADE)
awayteamID = models.ForeignKey(Team, null=False, on_delete=models.CASCADE)

making the below changes solved my issue:

hometeamID = models.ForeignKey(Team, null=False, on_delete=models.CASCADE,related_name='home_team')
awayteamID = models.ForeignKey(Team, null=False, on_delete=models.CASCADE,related_name='away_team')

Answer 3

This isn't an ultimate answer for the question, however for someone it may solve the problem. I got the same error in my project after checking out a really old commit (going to detached head state) and then getting the code base back up to date. Solution was to delete all *.pyc files in the project.

Answer 4

"If a model has a ForeignKey, instances of the foreign-key model will have access to a Manager that returns all instances of the first model. By default, this Manager is named FOO_set, where FOO is the source model name, lowercased."

But if you have more than one foreign key in a model, django is unable to generate unique names for foreign-key manager.
You can help out by adding "related_name" arguments to the foreignkey field definitions in your models.

See here: https://docs.djangoproject.com/en/dev/topics/db/queries/#following-relationships-backward

Answer 5

But in my case i am create a separate app for some functionality with same model name and field ( copy/paste ;) ) that's because of this type of error occurs i am just deleted the old model and code will work fine
May be help full for beginners like me :)

Answer 6

You have a number of foreign keys which django is unable to generate unique names for.

You can help out by adding "related_name" arguments to the foreignkey field definitions in your models. Eg:

content_type = ForeignKey(Topic, related_name='topic_content_type')

See here for more. http://docs.djangoproject.com/en/dev/ref/models/fields/#django.db.models.ForeignKey.related_name