1

api 链接不起作用。显示此错误“ init () 接受 1 个位置参数，但给出了 2 个”

请帮我。我是 django 的初学者

尝试创建博客 api model.py：

class Blog(models.Model):
    blog_id = models.AutoField(primary_key=True, editable=False)
    title = models.CharField(max_length=100, null=False)
    description = models.TextField(null=False)
    image = models.ImageField(null=True, blank=True)
    user = models.ForeignKey(User, on_delete=models.SET_NULL, null=True)
    created_at = models.DateTimeField(auto_now_add=True)
    updated_at = models.DateTimeField(auto_now=True)

    def __str__(self):
        return self.title

视图.py

class BlogViewSet(viewsets.ModelViewSet):
    serializer_class = BlogSerializer
    permission_classes = IsAuthenticated
    queryset = Blog.objects.all()

    def perform_create(self, serializer):
        serializer.save(user=self.request.user)

序列化程序.py：

class BlogSerializer(serializers.ModelSerializer):
    user = serializers.SerializerMethodField(read_only=True)

    class Meta:
        model = Blog
        fields = '__all__'

网址.py

from django.urls import path
from rest_framework.routers import DefaultRouter

from . import views
from rest_framework_simplejwt.views import (
    TokenObtainPairView,
)

router = DefaultRouter()
router.register('blogs', views.BlogViewSet, basename='blogs')

urlpatterns = [
    path('users/login', views.MyTokenObtainPairView.as_view()),
    path('users/register', views.register_user, name='register'),
    path('users/profile', views.get_user_profile, name='profile'),

    path('blogs/', views.BlogViewSet, name='blogs'),
    path('avatar/', views.get_avatar, name='avatar'),
]

因此，当我对“http://localhost:8000/api/blogs/”执行获取请求时，它会显示此错误：

TypeError at /api/blogs/

__init__() takes 1 positional argument but 2 were given

Request Method:     GET
Request URL:    http://localhost:8000/api/blogs/
Django Version:     3.2.4
Exception Type:     TypeError
Exception Value:    

__init__() takes 1 positional argument but 2 were given

Exception Location:     D:\work environment\Django_Api\codersavvy\lib\site-packages\django\core\handlers\base.py, line 181, in _get_response
Python Executable:  D:\work environment\Django_Api\codersavvy\Scripts\python.exe
Python Version:     3.9.2
Python Path:    

['D:\\work environment\\Django_Api\\codersavvy',
 'C:\\Users\\S\\AppData\\Local\\Programs\\Python\\Python39\\python39.zip',
 'C:\\Users\\S\\AppData\\Local\\Programs\\Python\\Python39\\DLLs',
 'C:\\Users\\S\\AppData\\Local\\Programs\\Python\\Python39\\lib',
 'C:\\Users\\S\\AppData\\Local\\Programs\\Python\\Python39',
 'D:\\work environment\\Django_Api\\codersavvy',
 'D:\\work environment\\Django_Api\\codersavvy\\lib\\site-packages']

Server time:    Mon, 14 Jun 2021 17:28:12 +0000
4

1 回答 1

1

您需要调用.as_view()将（基于类的）ViewSet转换为将调度 HTTP 调用的函数：

path('blogs/', views.BlogViewSet.as_view(), name='blogs'),

但是，由于您制作了，因此您可以将路由器[drf-doc]用于：ViewSet

from django.urls import path
from rest_framework.routers import DefaultRouter

from . import views
from rest_framework_simplejwt.views import (
    TokenObtainPairView,
)

router = DefaultRouter()
router.register('blogs', views.BlogViewSet, basename='blogs')

urlpatterns = [
    path('users/login', views.MyTokenObtainPairView.as_view()),
    path('users/register', views.register_user, name='register'),
    path('users/profile', views.get_user_profile, name='profile'),

    path('avatar/', views.get_avatar, name='avatar'),
]

urlpatterns += router.urls
于 2021-06-14T17:46:47.970 回答