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ViewsDjango, URL no encontrada 404
Estructura de la carpeta del proyecto
Soy nuevo en Django. Estoy tratando de crear un punto final de API de descanso simple, pero obtengo un 404. Estoy seguro de que me falta alguna configuración. modelos.py
class DisplayItems(models.Model): yesNo = models.CharField(max_length=10) yogaMessage = models.CharField(max_length=50) yogaTimeMessage = models.CharField(max_length=100) @classmethod def create(cls, yesNo, yogaMessage, yogaTimeMessage): displayItems = cls(yesNo=yesNo, yogaMessage=yogaMessage, yogaTimeMessage=yogaTimeMessage) return displayItemsserializadores.py
from .models import DisplayItems class DisplayItemsSerializer(serializers.ModelSerializer): yesNo = serializers.CharField(max_length=10) yogaMessage = serializers.CharField(max_length=50) yogaTimeMessage = serializers.CharField(max_length=100) class Meta: model = DisplayItems fields = ('__all__')urls.py
from .views import DisplayItemsViews from django.contrib import admin urlpatterns = [ path(r'^admin/', admin.site.urls), path(r'^zzz/', DisplayItemsViews.as_view()), ]vistas.py
from rest_framework.response import Response from rest_framework import status from .serializers import DisplayItemsSerializer from .models import DisplayItems class DisplayItemsViews(APIView): def get(self, request): displayItems = DisplayItems.create("YES!", "There's yoga today", "At 2:00 pm Eastern Time") serializer = DisplayItemsSerializer(displayItems) return Response({"status": "success", "data": serializer.data}, status=status.HTTP_200_OK)La URL que intento ejecutar es http://127.0.0.1:8000/zzz/
He leído casi todas las publicaciones relevantes de stackoverflow, pero parece que no puedo entender lo que me estoy perdiendo.
1 answers
Answer question0
Tres cosas:
1 - ¿Ha instalado rest_framework en sus INSTALLED_APPS de su settings.py ?
2 - Su path() en su urls.py no debe contener la parte r'^ , lo que significa que las rutas adecuadas son path('admin/', admin.site.urls) y path('zzz/', DisplayItemsViews.as_view()) . Este es un comportamiento antiguo de cuando Django sugirió usar url() en lugar de path()
3 - Supongo que ha importado APIView en sus views.py , si no, necesita from rest_framework.views import APIView
EDITAR: Mi versión local de trabajo.
vistas.py
from rest_framework.response import Response from rest_framework import status from rest_framework.views import APIView from .serializers import DisplayItemsSerializer from .models import DisplayItems class DisplayItemsViews(APIView): def get(self, request): displayItems = DisplayItems.create("YES!", "There's yoga today", "At 2:00 pm Eastern Time") serializer = DisplayItemsSerializer(displayItems) return Response({"status": "success", "data": serializer.data}, status=status.HTTP_200_OK)configuración.py
INSTALLED_APPS = [ 'django.contrib.admin', 'django.contrib.auth', 'django.contrib.contenttypes', 'django.contrib.sessions', 'django.contrib.messages', 'django.contrib.staticfiles', 'base', 'rest_framework' ]urls.py
from django.contrib import admin from django.urls import path from base.views import DisplayItemsViews urlpatterns = [ path('admin/', admin.site.urls), path('zzz/', DisplayItemsViews.as_view()) ]modelos.py
class DisplayItems(models.Model): yesNo = models.CharField(max_length=10) yogaMessage = models.CharField(max_length=50) yogaTimeMessage = models.CharField(max_length=100) @classmethod def create(cls, yesNo, yogaMessage, yogaTimeMessage): displayItems = cls(yesNo=yesNo, yogaMessage=yogaMessage, yogaTimeMessage=yogaTimeMessage) return displayItemsserializadores.py
from rest_framework import serializers from .models import DisplayItems class DisplayItemsSerializer(serializers.ModelSerializer): yesNo = serializers.CharField(max_length=10) yogaMessage = serializers.CharField(max_length=50) yogaTimeMessage = serializers.CharField(max_length=100) class Meta: model = DisplayItems fields = ('__all__') Y el resultado, en localhost:8080/zzz 