【发布时间】:2017-02-05 08:44:47
【问题描述】:
我正在使用 Django rest 框架,我创建了这个类来返回项目的所有名称
class cpuProjectsViewSet(viewsets.ViewSet):
serializer_class = serializers.cpuProjectsSerializer
def list(self, request):
all_rows = connect_database()
name_project = []
all_projects = []
for item_row in all_rows:
name_project.append(item_row['project'])
name_project = list(sorted(set(name_project)))
for i in range(0, len(name_project)):
all_projects.append({'project' : str(name_project[i])})
serializer = serializers.cpuProjectsSerializer(instance=all_projects, many=True)
return Response(serializer.data)
我的 URL 是这样的 http://127.0.0.1:8000/cpuProjects/ 这个返回所有项目,购买 如果现在我想要一个特定的项目,我是否要创建一个新类?如果我想使用相同的 URL ...例如
http://127.0.0.1:8000/cpuProjects/ => 返回所有项目
http://127.0.0.1:8000/cpuProjects/nameProject => 返回一个特定的项目。
class cpuProjectsViewSet(viewsets.ViewSet):
serializer_class = serializers.cpuProjectsSerializer
lookup_field = 'project_name'
def retrieve(self, request, project_name=None):
try:
opc = {'name_proj' : project_name }
all_rows = connect_database(opc)
except KeyError:
return Response(status=status.HTTP_404_NOT_FOUND)
except ValueError:
return Response(status=status.HTTP_400_BAD_REQUEST)
serializer = serializers.cpuProjectsSerializer(instance=all_rows, many=True)
return Response(serializer.data)
可以在同一个班级做吗?我尝试使用检索方法,但需要项目的 ID,没有名称对吗?
提前致谢!
【问题讨论】:
标签: python django django-rest-framework