Hye*_*eok 12 django django-models django-rest-framework
我是一名学习 django rest 框架的学生
我正在用 django rest 框架制作一个简单的 sns
我需要跟随系统。所以,我试图做到,但有一些麻烦
首先这是我的带有 AbstractBaseUser 和 PermissionsMixin 的用户模型
class User(AbstractBaseUser, PermissionsMixin):
user_id = models.CharField(max_length=100, unique=True, primary_key=True)
name = models.CharField(max_length=100)
created_at = models.DateTimeField(auto_now_add=True)
is_staff = models.BooleanField(default=False)
followers = models.ManyToManyField('self', related_name='follower',blank=True)
following = models.ManyToManyField('self', related_name='following',blank=True)
profile_image = models.ImageField(blank=True)
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现场追随者是追随我的人,追随的是我追随的人
当我使用此 APIView 类添加以下内容时
class AddFollower(APIView):
permission_classes = [IsAuthenticated, ]
def post(self, requset, format=None):
user = User.objects.get(user_id=self.request.data.get('user_id'))
follow = User.objects.get(user_id=self.request.data.get('follow'))
user.following.add(follow)
user.save()
follow.followers.add(user)
follow.save()
print(str(user) + ", " + str(follow))
return JsonResponse({'status':status.HTTP_200_OK, 'data':"", 'message':"follow"+str(follow.user_id)})
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user_id 是我,follow 是我想要关注的人
我想将关注添加到 user_id 的关注字段并将 user_id 添加到关注的关注字段
但它不起作用
我想要的结果是这样的(带有用户信息api)
{
"followers": [],
"following": [
"some user"
],
}
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一些用户的用户信息
{
"followers": [
"user above"
],
"following": [
],
}
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但真正的结果是这样的
{
"followers": [
"some user"
],
"following": [
"some user"
],
}
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一些用户的用户信息
{
"followers": [
"user above"
],
"following": [
"user above"
],
}
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这不是我想要的
我不知道这个问题我需要一些帮助
谢谢
Ent*_*tin 23
我会以不同的方式设计它。
我不会将信息添加到User模型中,而是明确创建另一个表来存储有关“关注者”和“关注者”的信息。
表的架构将是:
class UserFollowing(models.Model):
user_id = models.ForeignKey("User", related_name="following")
following_user_id = models.ForeignKey("User", related_name="followers")
# You can even add info about when user started following
created = models.DateTimeField(auto_now_add=True)
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现在,在您的 post 方法实现中,您只需执行以下操作:
UserFollowing.objects.create(user_id=user.id,
following_user_id=follow.id)
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然后,您可以轻松访问以下和关注者:
user = User.objects.get(id=1) # it is just example with id 1
user.following.all()
user.followers.all()
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然后您可以创建约束,这样用户就不能两次关注同一个用户。但我把这留给你(提示:unique_together)
uni*_*dev 13
上述解决方案很好,而且是最佳的,但我想为任何想要实现此类功能的人提供详细的解决方案。
from django.contrib.auth import get_user_model
UserModel = get_user_model()
class UserFollowing(models.Model):
user_id = models.ForeignKey(UserModel, related_name="following", on_delete=models.CASCADE)
following_user_id = models.ForeignKey(UserModel, related_name="followers", on_delete=models.CASCADE)
created = models.DateTimeField(auto_now_add=True, db_index=True)
class Meta:
constraints = [
models.UniqueConstraint(fields=['user_id','following_user_id'], name="unique_followers")
]
ordering = ["-created"]
def __str__(self):
f"{self.user_id} follows {self.following_user_id}"
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class UserFollowingViewSet(viewsets.ModelViewSet):
permission_classes = (IsAuthenticatedOrReadOnly,)
serializer_class = UserFollowingSerializer
queryset = models.UserFollowing.objects.all()
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class FollowingSerializer(serializers.ModelSerializer):
class Meta:
model = UserFollowing
fields = ("id", "following_user_id", "created")
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class FollowersSerializer(serializers.ModelSerializer):
class Meta:
model = UserFollowing
fields = ("id", "user_id", "created")
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from django.contrib.auth import get_user_model
User = get_user_model()
class UserSerializer(serializers.ModelSerializer):
following = serializers.SerializerMethodField()
followers = serializers.SerializerMethodField()
class Meta:
model = User
fields = (
"id",
"email",
"username",
"following",
"followers",
)
extra_kwargs = {"password": {"write_only": True}}
def get_following(self, obj):
return FollowingSerializer(obj.following.all(), many=True).data
def get_followers(self, obj):
return FollowersSerializer(obj.followers.all(), many=True).data
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小智 5
models.py
from django.contrib.auth.models import AbstractUser
class User(AbstractUser):
followers = models.ManyToManyField('self', symmetrical=False,
blank=True)
def count_followers(self):
return self.followers.count()
def count_following(self):
return User.objects.filter(followers=self).count()
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