Sql Django admin：以编程方式选择内联中的可见字段集

我想要的是能够根据django Admin中ForeignKey选择器的选择来选择要显示的表单集

我有四种型号：

产品

评级

准则（问题）

标准功率

我的数据库组织如下：

Product <- ManyToOne -- Rating <- ManyToMany -> Criterion -
    |                                                  |
    -- OneToMany ->   CriterionAnswer    <- OneToMany --

################
# Product
################
class Product (models.Model):

rating = models.ForeignKey(
    'Rating',
    on_delete=models.SET_NULL,
    null=True,
)

def save(self, *args, **kwargs):

    super(Product, self).save(*args, **kwargs)

    # get all criteria that need an answer in order to create the rating
    criteriaToAnswer = Criterion.objects.filter(rating=self.rating)
    # iterate over all criterias and check if the answer exists in database
    for crit in criteriaToAnswer:
        # if it doesn't exist, create an answer
        if not CriterionAnswer.objects.filter(criterion=crit).filter(product=self):
            print "Criteria answer to: <%s> does not exist... creating new one" % crit.__str__()
            new_criterionAnswer = CriterionAnswer()
            new_criterionAnswer.criterion = crit
            new_criterionAnswer.product = self
            new_criterionAnswer.save()
        else:
            print "Criteria answer to: <%s> exists." % crit.__str__()

################
# CriterionAnswer
################
class CriterionAnswer(models.Model):

criterion = models.ForeignKey('Criterion', on_delete=models.CASCADE)
product = models.ForeignKey('Product', on_delete=models.CASCADE)

################
# Rating
################

class Rating(models.Model):

name = models.CharField(max_length=30)

################
# Criterion
################

class Criterion(models.Model):

category = models.ForeignKey(
    'Category',
    on_delete=models.PROTECT,
)

rating = models.ManyToManyField(
    'Rating',
    through='CriterionInRating',
    through_fields=('criterion', 'rating'),
)

class ProductAdmin(admin.ModelAdmin):

inlines = [
    # Something here...
]

# Or something here that could solve my problem.