如何立即更改 Django 模型中对象的记录？

Question

我有一个 class 及其对象，我想在保存新记录时，Django 检查 "is_it_true" 是 False 还是 True，如果是 True 则它会更改 "number" 到一个新值，该值对应于具有 "is_it_question" 加一的对象的计数；如果它是 false，"number" 的默认值将保留为默认值 (100)。我该怎么办？

class TypeFourChoice(models.Model):
    question_choice = models.ForeignKey(TypeFour)
    is_it_question = models.BooleanField(default=False)
    number = models.PositiveIntegerField(blank=False, default=100)
    word_or_words = models.CharField(default='', blank=False, max_length=20)
    timestamp = models.DateTimeField(auto_now_add=True)

    def __str__(self):
        return "{}".format(self.question_choice)

    @classmethod
    def save(cls, *args, **kwargs):
        no = cls.objects.exclude(number=100).count()
        if no is None:
            pass
        else:
            clss = cls.objects.get(???????)
            if clss.is_it_question:
                no += 1
                clss.number = no
                clss.save(*args, **kwargs)
            else:
                clss.number = 100
                clss.save(*args, **kwargs)

Answer 1

您应该从 save() 方法中删除 @classmethod 装饰器。并将方法简化为：

def save(self, *args, **kwargs):
    # New objects have no `pk`
    if self.pk is None and self.is_it_question:
        self.number = TypeFourChoice.objects.filter(is_it_question=True) \
                                            .count() + 1
    super(TypeFourChoice, self).save(*args, **kwargs)

如何立即更改 Django 模型中对象的记录？

How to change record of object in Django's model instantly?

python

django

django-models

django-admin