我需要扩展用户模型以添加地址,分数,更多user_types等内容。有两种可能的实现方式,扩展用户模型或创建一个新模型,该模型将通过{ {1}}。我决定使用新模型,因为它看起来更容易,并且在this堆栈溢出问题中推荐使用。但是,现在我无法创建没有嵌套配置文件字段的序列化程序,而该字段未记录,因为默认的rest_framwork文档生成器无法生成嵌套序列化程序的文档。
我的OneToOneField
如下:
UserSerializer
此序列化器采用以下JSON格式:
class UserSerializer(serializers.ModelSerializer):
# This creates a nested profile field
profile = ProfileSerializer(required=True)
def create(self, validated_data):
profile_data = validated_data.pop('profile')
user = User.objects.create_user(**validate_data)
profile, created = Profile.objects.upodate_or_creeate(user=user, defaults=profile_data)
return user
class Meta:
model = User
fields = ('id', 'username', 'email', 'password', 'buckelists', 'profile')
read_only_fields = ('id',)
extra_kwargs = {'password':{'write_only': True}}
这看起来很奇怪,用{
'name': ...,
'email': ...,
'password': ...,
'profile': {
'address': ...,
'score': ...,
'user_type': ...,
'achievements': ...,
'country': ...,
'trusted': ...,
}
生成的文档显示如下:
rest_framework.documentation.include_docs_urls
因此,不清楚配置文件字段中应包含哪些内容。我想创建将接受以下格式的序列化器:
{
'username': ...,
'email': ...,
'password': ...,
'field': ...,
}
是否可以从头开始创建自定义序列化程序?或者至少可以为嵌套序列化程序生成文档。
PS:我使用python3.6和Django 2.1
编辑: 这是我的models.py的相关部分:
{
'name': ...,
'email': ...,
'password': ...,
'address': ...,
'score': ...,
'user_type': ...,
'achievements': ...,
'country': ...,
'trusted': ...,
}
编辑:
Mohammad Ali的答案为GET解决了这个问题,但是我也想使用POST,UPDATE和PATCH方法。我发现我必须使用class Profile(models.Model):
user = models.OneToOneField(User, on_delete=models.CASCADE)
trusted = models.BooleanField(default=False)
address = models.CharField(max_length=100, default="")
COUNTRIES = (
('CZ', 'Czech Republic'),
('EN', 'England'),
)
country = models.CharField(max_length=2, choices=COUNTRIES, default="CZ")
score = models.BigIntegerField(default=0)
achievements = models.ManyToManyField(Achievement, blank=True)
USER_TYPES = (
('N', 'Normal'),
('C', 'Contributor'),
('A', 'Admin'),
)
user_type = models.CharField(max_length=1, choices=USER_TYPES, default='N')
@receiver(post_save, sender=settings.AUTH_USER_MODEL)
def create_auth_token(sender, instance=None, created=False, **kwargs):
if created:
Token.objects.create(user=instance)
@receiver(post_save, sender=User)
def create_user_profile(sender, instance, created=False, **kwargs):
if created:
profile, created = Profile.objects.get_or_create(user=instance)
profile.save()
参数,但这与序列化程序有关,我不知道如何在没有配置文件字段的情况下引用配置文件。
答案 0 :(得分:1)
放轻松,您可以仅在create函数中创建Profile obj。
class UserSerializer(serializers.ModelSerializer):
trusted = serializers.BooleanField()
address = serializers.CharField()
class Meta:
model = User
fields = ('username', 'email', 'password', 'trusted', 'address',)
def create(self, validated_data):
user = User.objects.create(username=validated_data['username'], email=validated_data['email'])
user.set_password(validated_data['password'])
user.save()
profile = Profile(user=user, trusted=validated_data['trusted'], address=validated_data['address']
profile.save()
return validated_data
这只是您的方案的简短实现。您可以填充故事。
答案 1 :(得分:-1)
请阅读有关序列化程序的文档:Django REST FRAMEWORK -用户related_name
user = models.OneToOneField(User, on_delete=models.CASCADE, related_name="user_profile") # models
class ProfileSerializer(serializers.ModelSerializer):
user = serializers.PrimaryKeyRelatedField(queryset=User.objects.all(), required=False)
class Meta:
model = Profile
fields = '__all__'
class UserSerializer(serializers.ModelSerializer):
user_profile = ProfileSerializer(required=True)
class Meta:
model = User
fields = '__all__'