python - 如何在 Django 中实现地址表单？这就是我所拥有的

Question

大家好，我正在尝试弄清楚如何向用户添加多个地址。所以用户可以有送货地址和家庭地址。我有点猜到四处阅读，但它不起作用。

我还创建了一个简单的模式(我忘了包括邮政编码):

模型.py

class Address(models.Model):
    name = models.CharField(max_length=30)
    address = models.CharField(max_length=50)
    city = models.CharField(max_length=60, default="Miami")
    state = models.CharField(max_length=30, default="Florida")
    zipcode = models.CharField(max_length=5, default="33165")
    country = models.CharField(max_length=50)

    class Meta:
        verbose_name = 'Address'
        verbose_name_plural = 'Address'

    def __str__(self):
        return self.name

# All user data is/should be linked to this profile, so when user gets deleted, all data deletes as well
class Profile(models.Model):
    user = models.OneToOneField(User, on_delete=models.CASCADE)
    nick_name = models.CharField('Nick name', max_length=30, blank=True, default='')
    bio = models.TextField(max_length=500, blank=True)
    image = models.ImageField(default='default.jpg', upload_to='profile_pics')
    addresses = models.ManyToManyField(
        Address,
        through='AddressType',
        through_fields=('address', 'profile'),
    )

    # If we don't have this, it's going to say profile object only
    def __str__(self):
         return f'{self.user.username} Profile'  # it's going to print username Profile

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

            img = Image.open(self.image.path)

            if img.height > 300 or img.width > 300:
                output_size = (300, 300)
                img.thumbnail(output_size)
                img.save(self.image.path)


class AddressType(models.Model):

    HOME_ADDRESS = 1
    SHIPPING_ADDRESS = 2

    TYPE_ADDRESS_CHOICES = (
        (HOME_ADDRESS, "Home address"),
        (SHIPPING_ADDRESS, "Shipping address"),
    )

    address = models.ForeignKey('Address', on_delete=models.CASCADE)
    profile = models.ForeignKey('Profile', on_delete=models.CASCADE)

    # This is the field you would use for know the type of address.
    address_type = models.PositiveIntegerField(choices=TYPE_ADDRESS_CHOICES)

当我执行 makemigrations 时，它说:

ERRORS:
users.Profile.addresses: (fields.E339) 'AddressType.address' is not a foreign key to 'Profile'.
        HINT: Did you mean one of the following foreign keys to 'Profile': profile?
users.Profile.addresses: (fields.E339) 'AddressType.profile' is not a foreign key to 'Address'.
        HINT: Did you mean one of the following foreign keys to 'Address': address?

有人可以帮我一下吗？

非常感谢

Answer 1

首先对您的设计发表评论...

因此一个用户可以有多个地址，区别在于可以是家庭或地址或送货地址。

你可以使用 ManyToManyField并通过第三个模型“描述”这种关系，该模型将包含运送信息或家庭地址信息。

首先，我会将您的“HomeAddress”重命名为“Address”，这样更加语义化，然后使用 through 与第三个表建立关系。

阅读ManyToManyFiled文档以获取更多详细信息。

例子:

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

class Profile(models.Model):
    addresses = models.ManyToManyField(
         'Address', 
         through='AddressInfo'
         through_fields=('address', 'profile')
    )
    # ...

class AddressInfo(models.Model):

    HOME_ADDRESS = 1
    SHIPPING_ADDRESS = 2

    TYPE_ADDRESS_CHOICES = (
        (HOME_ADDRESS, "Home address"),
        (SHIPPIN_ADDRESS, "Shipping address"),
    )

    address = models.ForeignKey('Address', on_delete=models.CASCADE)
    profile = models.ForeignKey('Profile', on_delete=models.CASCADE)

    # This is the field you would use for know the type of address.
    address_type = models.PositiveIntegerField(choices=TYPE_ADDRESS_CHOICES)

关于创建表单...

然后您可以编写用于将地址添加到某些配置文件并考虑地址类型的表单。

如果要同时添加多个地址，建议使用FormSet或 ModelFormSet .