Django - filtering on foreign key properties

Question

I'm trying to filter a table in Django based on the value of a particular field of a foreign key.

For example I have two models -

# models.py
class Asset(models.Model):
 name = models.TextField(max_length=150)
 project = models.ForeignKey('Project')

class Project(models.Model):
 name = models.TextField(max_length=150)

I'd like to filter my asset list based on the name of the associated project.

Currently I am performing two queries:

# views.py
project_list = Project.objects.filter(name__contains="Foo")         
asset_list = Asset.objects.filter( desc__contains=filter, project__in=project_list).order_by('desc')

I'm wondering if there is a way to specify this kind of filtering in the main query?

Answer 1

student_user = User.objects.get(id=user_id)
available_subjects = Subject.objects.exclude(subject_grade__student__user=student_user) # My ans
enrolled_subjects = SubjectGrade.objects.filter(student__user=student_user)
context.update({'available_subjects': available_subjects, 'student_user': student_user, 
                'request':request, 'enrolled_subjects': enrolled_subjects})

In my application above, i assume that once a student is enrolled, a subject SubjectGrade instance will be created that contains the subject enrolled and the student himself/herself.

Subject and Student User model is a Foreign Key to the SubjectGrade Model.

In "available_subjects", i excluded all the subjects that are already enrolled by the current student_user by checking all subjectgrade instance that has "student" attribute as the current student_user

PS. Apologies in Advance if you can't still understand because of my explanation. This is the best explanation i Can Provide. Thank you so much

Answer 2

Asset.objects.filter( project__name__contains="Foo" )

Answer 3

This has been possible since the queryset-refactor branch landed pre-1.0. Ticket 4088 exposed the problem. This should work:

Asset.objects.filter(
    desc__contains=filter,
    project__name__contains="Foo").order_by("desc")

The Django Many-to-one documentation has this and other examples of following Foreign Keys using the Model API.