在Django应用程序中显示特定用户详细信息，而不是所有详细信息
在Django应用程序中显示特定用户详细信息，而不是所有详细信息
提问于 2018-12-16 21:15:18
我正在创建我的第一个Django项目。我接受了30,000个值作为输入，并希望根据主键显示特定的值。
代码：
class employeesList(APIView):
def get(self,request):
employees1 = employees.objects.all()
with open('tracking_ids.csv') as f:
reader = csv.reader(f)
for row in reader:
_, created = employees.objects.get_or_create(
AWB=row[0],
Pickup_Pincode=row[1],
Drop_Pincode=row[2],
)
serializer = employeesSerializer(employees1 , many=True)
return Response(serializer.data)
def post(self,request):
# employees1 = employees.objects.get(id=request.employees.AWB)
employees1 = employees.objects.all()
serializer = employeesSerializer(employees1 , many=True)
return Response(serializer.data) 如果我在URL中输入http://127.0.0.1:8000/employees/，我会得到所有的值。我希望网址类似于http://127.0.0.1:8000/employees/P01001168074，并显示P01001168074的值，其中P01001168074是主ID。
我读过1:showing the model values of specific user django 2)editing user details in python django rest framework，但它们不同
它能做到吗?如果它能做到，那么如何做到呢？
回答 2
Stack Overflow用户
回答已采纳
发布于 2018-12-16 21:40:49
假设您使用的是Django2.0，那么您必须配置一个能够捕获参数as seen in the documentation的路径
from rest_framework.views import APIView
from rest_framework.response import Response
from rest_framework.serializers import ModelSerializer
class A(models.Model):
x = models.CharField(max_length=250)
class MySerializer(ModelSerializer):
class Meta:
model = A
fields = ('x',)
class MyListView(APIView):
def get(self, request, *args, **kwargs):
# simply get all the objects, serialize and return the response
elems = A.objects.all()
response_data = MySerializer(elems, many=True).data
return Response(data=response_data)
class MyDetailView(APIView):
def get(self, request, *args, **kwargs):
# we obtain the parameter from the URL
desired_item = kwargs.get("desired_item", None)
# we filter the objects by it and get our instance
element = A.objects.filter(x=desired_item).first()
# serialize the instance and return the response
response_data = MySerializer(element).data
return Response(data=response_data)
# all we have to now is define the paths for the list and the detail views.
urlpatterns = [
path('employees/', MyListView.as_view()),
path('employees/<str:desired_item>', MyDetailView.as_view())
]Stack Overflow用户
发布于 2018-12-17 04:31:21
一个不错的选择是使用已经包含list和detail端点的viewset，并且对于默认的简单设置只需要很少的代码
views.py
from rest_framework import viewsets
class EmployeeViewSet(viewsets.ModelViewSet):
serializer_class = EmployeeSerializer
queryset = Employee.objects.all()urls.py
from rest_framework.routers import SimpleRouter
from views import EmployeeViewSet
router = SimpleRouter()
router.register(r'employees', EmployeeViewSet, base_name='employees')
urlpatterns = router.get_urls()您可以在DRF docs中阅读有关视图集的更多信息
页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接：
https://stackoverflow.com/questions/53802513
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