Question

大家好，我想为DRF中的帖子创建一个喜欢的功能

Creating a "like" url/button for Django Rest Framework

我从堆栈溢出中遵循了上述问题

这是我的模型。py

class status(models.Model):
    user=models.ForeignKey(settings.AUTH_USER_MODEL,on_delete=models.CASCADE, null=False)
    # user=models.ForeignKey(settings.AUTH_USER_MODEL,on_delete=models.CASCADE)

    content=models.TextField(blank=True,null=True)

    image=models.ImageField(upload_to="upload",null=True,blank=True)
    updated=models.DateTimeField(auto_now=True)
    time=models.DateTimeField(auto_now_add=True)
    likes=models.ManyToManyField(settings.AUTH_USER_MODEL,blank=True,related_name='post_likes')

    def __str__(self):
        return self.content

serializers.py

class statusSerializer(serializers.ModelSerializer):
    # user_name = serializers.ReadOnlyField(source='user.username')
    user=UserPublicSerializer(read_only=True)
    # rating=RatingSerializer(required=False)
    # rating=RatingSerializer()

    class Meta:
        model=status
        fields=['id','user','content','image','likes']

views.py

@api_view(['POST'])
def LikesApi(self,request,id):
    # status = get_object_or_404(id=request.POST.get('id', ''))
    status = get_object_or_404(id=id)

    status.likes.add(request.user)
    serializer = statusSerializer(status)
    return Response(serializer.data, status=status.HTTP_201_CREATED)

我已经尝试过从堆栈上考虑该问题的两种方式

@api_view(['POST'])
def LikesApi(self,request,id):
    serializer=statusSerializer(data=request.DATA)
    if serializer.is_valid():
        serializer.object.content_object=get_object_or_404(status,id=request.POST.get('id', ''))
        serializer.object.likes.add(request.user)
        serializer.save()
        return  RestResponse(serializer.data, status=status.HTTP_201_CREATED)
    return RestResponse(serializer.errors, status=status.HTTP_400_BAD_REQUEST)

这是我的urls.py

urlpatterns = [
        # url(r'^$/', StatusListSearchApi.as_view()),
        url(r'^list/', StatusListSearchApi.as_view()),
        url(r'^user/', UserPost.as_view()),

        url(r'^like/(?P<id>\d+)/$',LikesApi),

post对象存在于我的api / list / url

http://127.0.0.1:8000/api/list/like?id=35

当我触发此网址时，我无法在可浏览的api detailview中看到喜欢的增加

http://127.0.0.1:8000/api/like?id=35

也尝试过此操作，但页面未找到404错误

我不知道问题出在哪里，我该如何通过网址获取特定帖子的ID并处理视图

Answer 1

我建议您使其成为RESTful，并使其网址类似：/api/status/<status_id>/like，还可以查看naming conventions并为您修改姓名。

对于视图，这将是处理的好方法：

from django.shortcuts import get_object_or_404
from rest_framework.response import Response
from <your_package>.models import YourStatusModel

@api_view(['POST'])
def like_api_view(self, request, status_id):
    obj = get_object_or_404(YourStatusModel.objects.all(), id=status_id)
    obj.likes.add(request.user)

    return Response(status=status.HTTP_201_CREATED)

django：视图未从URL DJANGO REST FRAMEWORK中获取ID

1 个答案: