如何在 django 中实现地址表单？这就是我所拥有的

Question

大家好，我正在研究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。我有点猜想阅读，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）：

models.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)

当我进行迁移时，它会说：

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?

有人可以帮帮我吗？

非常感谢

Answer 1

首先对您的设计发表评论...

所以一个用户可以有许多个地址，区别在于可以是家庭地址或地址或送货地址。

您可以使用ManyToManyField 并通过第三个模型“描述”这种关系，该模型将包含运输或家庭地址的信息。

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

阅读ManyToManyFiled 文档了解更多详情。

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，您可以编写表单以将地址添加到某些配置文件，同时考虑地址类型。

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