django - 创建实例时如何更新 ManyToManyField

Question

我想在创建实例时更新字段，我尝试过 signals但对 ManyToManyField 来说似乎很复杂

class MobileCustomer(models.Model):
    customer = models.CharField(max_length=30)
    phone = models.CharField(max_length=13)
    mobile = models.ManyToManyField(MobileStorage,through='SelectMobile')

class SelectMobile(models.Model):
    mobile = models.ForeignKey(MobileStorage,on_delete=models.CASCADE)
    item = models.ForeignKey(MobileCustomer,on_delete=models.CASCADE)
    quantity = models.PositiveIntegerField(default=1)
    imei = models.ManyToManyField(Imei)

class MobileStorage(models.Model):
    mobile = models.ForeignKey(Mobile,on_delete=models.CASCADE)
    quantity = models.PositiveIntegerField()

class Mobile(models.Model):
    mobile_name = models.CharField(max_length=40,unique=True)

class Imei(models.Model):
    imei = models.CharField(max_length=15,verbose_name='IMEI',unique=True)
    mobile = models.ForeignKey(MobileStorage,on_delete=models.CASCADE)
    active = models.BooleanField(default=True)

我想更新 Imei 中的事件字段型号当 MobileCustomer创建!？
我在我的 CreateView form_valid 中试过这个

imei = Imei.objects.filter(selectmobile__item=self.object).update(active=False)
print(imei)

它打印了 0 ？
每当在 createview 中创建实例时，如何使用信号或 update() 使其工作

Answer 1

查找问题。 mobile 查找到 MobileStorage，然后 mobilecustom 查找到 MobileCustomer。

def form_valid(self, form):
  # self.object = None
  super().form_valid(form)
  # self.object = a instance of MobileCustomer
  imei = Imei.objects.filter(mobile__mobilecustomer=self.object).update(active=False)
  print(imei)