我有一个非常基本的Django基于类的视图如下。它甚至没有做任何事情:
from django.views.generic import View, TemplateView
class index(TemplateView):
template_name = 'myApp/index.html'
def dispatch(self, request, *args, **kwargs):
return super(index, self).dispatch(request, *args, **kwargs)
但是,当我通过浏览器加载此视图时,出现以下错误:
Exception Type: TypeError
Exception Value:
__init__() takes exactly 1 argument (2 given)
Exception Location: /Library/Python/2.7/site-packages/Django-1.6.5-py2.7.egg/django/core/handlers/base.py in get_response, line 112
Python Executable: /usr/bin/python
Python Version: 2.7.5
Python Path:
为什么我收到此错误?以及如何解决它?
下面是关于错误位置的更多背景信息:
/Library/Python/2.7/site-packages/Django-1.6.5-py2.7.egg/django/core/handlers/base.py in get_response
106. response = middleware_method(request, callback, callback_args, callback_kwargs)
107. if response:
108. break
109. if response is None:
110. wrapped_callback = self.make_view_atomic(callback)
111. try:
112. response = wrapped_callback(request, *callback_args, **callback_kwargs) ...
113. except Exception as e:
114. # If the view raised an exception, run it through exception
115. # middleware, and if the exception middleware returns a
116. # response, use that. Otherwise, reraise the exception.
117. for middleware_method in self._exception_middleware:
118. response = middleware_method(request, e)