我已经将swagger与django rest框架集成在一起,但是swagger docs没有创建一个输入框来发布post请求的数据 .
我的解析器设置,
REST_FRAMEWORK = {
'DEFAULT_PARSER_CLASSES': (
'rest_framework.parsers.JSONParser',
'rest_framework.parsers.FormParser',
'rest_framework.parsers.MultiPartParser',
),
}
这是我的视图类片段,`
class TeamViewList(APIView, BaseView):
"""
Class based view to handle all operations related to Team Model
"""
logger = logging.getLogger(__name__)
def get_serializer(self):
return serializers.TeamSerializer
def post(self, request):
"""
To create a new team
"""
try:
urlMapping中:
urlpatterns = [
url(r'^role/$', rest_views.UserTeamRoleView.as_view(), name='user_team_role'),
url(r'^teams/$', rest_views.TeamViewList.as_view(), name='team_list'),
url(r'^teams/(?P<name>[_a-zA-Z0-9\-]+)$', rest_views.TeamViewDetail.as_view(), name='team_detail'),
]
我的招摇文档生成,
无法将json有效负载作为输入传递给post请求 .
2 回答
您必须为视图选择
PARSER
或将DEFAULT_PARSER_CLASSES
添加到DRF设置以在Swagger上查看表单或json对象 .以下方法应返回实例而不是类:
def get_serializer(self): return serializers.TeamSerializer
顺便说一下,即使使用实例而不是类,它在DRF
3.6.2
上也失败了 . 但是,它看起来像has been fixed here