当在 django url 中使用 include 时，Django Swagger 开始失败

Question

我正在为我的 django 项目使用 django rest_framework_swagger，一切正常，但是当我使用 include 方法添加一些 URL 时，Swagger 开始给我 500 internal server error.

我不确定为什么会出现此错误，我已经检查但没有找到任何可以修复此错误的方法。

我正在使用：

django 1.11.7
rest_framework_swagger 2.1.2
django rest framework 3.7.3

网址

from django.conf.urls import url, include
from link_one.views import LinkOneViewSet
from link_two.views import LinkTwoViewSet

schema_view = get_swagger_view(title='My Project APIs')

urlpatterns = [

    url(r'^$', schema_view),
    url(r'^foo/(?P<foo_id>\w+)/bar/(?P<bar_id>\w+)/link1',
        LinkOneViewSet.as_view({'get': 'list'})),
    url(r'^foo/(?P<foo_id>\w+)/bar/(?P<bar_id>\w+)/link2',
        LinkTwoViewSet.as_view({'get': 'list'})),
    url(r'^foo/(?P<foo_id>\w+)/bar/(?P<bar_id>\w+)/link3',
        include('link_three.urls'))
]+ static(settings.STATIC_URL, document_root=settings.STATIC_ROOT)

错误

    [25/Jan/2021 14:03:31] ERROR [django.request.exception:135] Internal Server Error: /
Traceback (most recent call last):
  File "C:\Users\myuser\conda_env\lib\site-packages\django\core\handlers\exception.py", line 41, in inner
    response = get_response(request)
  File "C:\Users\myuser\conda_env\lib\site-packages\django\core\handlers\base.py", line 187, in _get_response
    response = self.process_exception_by_middleware(e, request)
  File "C:\Users\myuser\conda_env\lib\site-packages\django\core\handlers\base.py", line 185, in _get_response
    response = wrapped_callback(request, *callback_args, **callback_kwargs)
  File "C:\Users\myuser\conda_env\lib\site-packages\django\views\decorators\csrf.py", line 58, in wrapped_view
    return view_func(*args, **kwargs)
  File "C:\Users\myuser\conda_env\lib\site-packages\django\views\generic\base.py", line 68, in view
    return self.dispatch(request, *args, **kwargs)
  File "C:\Users\myuser\conda_env\lib\site-packages\rest_framework\views.py", line 489, in dispatch
    response = self.handle_exception(exc)
  File "C:\Users\myuser\conda_env\lib\site-packages\rest_framework\views.py", line 449, in handle_exception
    self.raise_uncaught_exception(exc)
  File "C:\Users\myuser\conda_env\lib\site-packages\rest_framework\views.py", line 486, in dispatch
    response = handler(request, *args, **kwargs)
  File "C:\Users\myuser\conda_env\lib\site-packages\rest_framework_swagger\views.py", line 32, in get
    schema = generator.get_schema(request=request)
  File "C:\Users\myuser\conda_env\lib\site-packages\rest_framework\schemas\generators.py", line 278, in get_schema
    links = self.get_links(None if public else request)
  File "C:\Users\myuser\conda_env\lib\site-packages\rest_framework\schemas\generators.py", line 316, in get_links
    link = view.schema.get_link(path, method, base_url=self.url)
  File "C:\Users\myuser\conda_env\lib\site-packages\rest_framework\schemas\inspectors.py", line 179, in get_link
    fields += self.get_serializer_fields(path, method)
  File "C:\Users\myuser\conda_env\lib\site-packages\rest_framework\schemas\inspectors.py", line 302, in get_serializer_fields
    serializer = view.get_serializer()
  File "C:\Users\myuser\conda_env\lib\site-packages\rest_framework\generics.py", line 112, in get_serializer
    return serializer_class(*args, **kwargs)
TypeError: 'list' object is not callable

Answer 1

经过大量的搜索和调试，我找到了解决这个问题的方法。

解决方案是，不要对 ViewSet 使用多个序列化器 classes。

在我的一个视图中，我正在这样做，这就是造成问题的原因。

class FooBarViewset(ModelViewSet): 
    serializer_class = [DefaultSerializer, BarSerializer, FooSerializer]

但是我没有意识到这会导致错误。

这是我正在使用的 Fix

class FooBarViewset(ModelViewSet): 
    serializer_class = DefaultSerializer

您还可以使用 get_serializer 方法和地图序列化程序 class 进行操作，请查看此答案 Django rest framework, use different serializers in the same ModelViewSet

当在 django url 中使用 include 时，Django Swagger 开始失败

Django Swagger starts failing when include is used in django urls

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