django - 使用两个可选但一个必需的外键创建模型

Question

我的问题是我有一个模型，可以使用两个外键之一来说明它是哪种模型。我希望它至少要拿一个，但不能两个都拿。我可以让它仍然是一个模型还是应该将其分为两种类型。这是代码:

class Inspection(models.Model):
    InspectionID = models.AutoField(primary_key=True, unique=True)
    GroupID = models.ForeignKey('PartGroup', on_delete=models.CASCADE, null=True, unique=True)
    SiteID = models.ForeignKey('Site', on_delete=models.CASCADE, null=True, unique=True)

    @classmethod
    def create(cls, groupid, siteid):
        inspection = cls(GroupID = groupid, SiteID = siteid)
        return inspection

    def __str__(self):
        return str(self.InspectionID)

class InspectionReport(models.Model):
    ReportID = models.AutoField(primary_key=True, unique=True)
    InspectionID = models.ForeignKey('Inspection', on_delete=models.CASCADE, null=True)
    Date = models.DateField(auto_now=False, auto_now_add=False, null=True)
    Comment = models.CharField(max_length=255, blank=True)
    Signature = models.CharField(max_length=255, blank=True)

问题是 Inspection模型。这应该链接到组或站点，但不能同时链接到两者。目前，通过此设置，它同时需要两者。

我宁愿不必将其分为两个几乎相同的模型 GroupInspection和 SiteInspection，因此将其保持为一个模型的任何解决方案都是理想的。

Answer 1

我建议您对Django way进行此类验证

通过覆盖Django模型的clean方法

class Inspection(models.Model):
    ...

    def clean(self):
        if <<<your condition>>>:
            raise ValidationError({
                    '<<<field_name>>>': _('Reason for validation error...etc'),
                })
        ...
    ...

Note, however, that like Model.full_clean(), a model’s clean() method is not invoked when you call your model’s save() method. it needs to be called manually to validate model's data, or you can override model's save method to make it always call the clean() method before triggering the Model class save method

Another solution that might help is using GenericRelations, in order to provide a polymorphic field that relates with more than one table, but it can be the case if these tables/objects can be used interchangeably in the system design from the first place.