具有用户定义字段和外键选项的 ModelForm

问：

在我的应用程序中，我有作为中心模型。s 有多个 ，每个有多个 s。用户通过为链接到该研究的每个级别选择一个级别来为该研究创建 s，例如：StudyStudyStrataStrataLevelAllocationStrata

class Study(models.Model):
    name = models.CharField(max_length=100)

class Stratum(models.Model):
    study = models.ForeignKey(Study, on_delete=models.CASCADE, related_name='strata')
    name = models.CharField(max_length=100)

class Level(models.Model):
    stratum = models.ForeignKey(Stratum, on_delete=models.CASCADE, related_name='levels')
    label = models.CharField(max_length=100)

class Allocation(models.Model):
    study = models.ForeignKey(Study, on_delete=models.CASCADE, related_name='allocations')
    code = models.CharField(blank=False, max_length=100)
    levels = models.ManyToManyField(Level, related_name='allocations')

为了为分配创建表单创建字段，我目前正在表单的构造函数中找到所有和关联的级别，但隐藏了层，因为用户不与它们交互：Strata

class AllocationForm(forms.ModelForm):

    class Meta:
        model = Allocation
        fields = ('code',)

    def __init__(self, *args, **kwargs):
        study = kwargs.pop('study')
        super(AllocationForm, self).__init__(*args, **kwargs)

        strata = Stratum.objects.filter(study=study)

        for stratum in strata:
            self.fields[stratum.name] = forms.IntegerField(
                widget=forms.HiddenInput()
            )
            self.fields[stratum.name].initial = stratum.id
            self.fields[stratum.name].disabled = True

            self.fields[stratum.name + '_level'] = forms.ModelChoiceField(
                queryset=Level.objects.filter(stratum=stratum)
            )

这是将关联对象附加到表单的安全且明智的方法吗？我担心在尝试创建分配时会忘记 和 s 之间的联系。这在视图中会更好吗？StrataLevel