我在Pyramid中创建应用时遇到了问题。当我尝试通过贴纸服务时,我得到:
File "/home/viraptor/blah/blah/__init__.py", line 23, in main
return config.make_wsgi_app()
File "/home/viraptor/pyramid/lib/python2.6/site-packages/pyramid/config.py", line 916, in make_wsgi_app
self.commit()
File "/home/viraptor/pyramid/lib/python2.6/site-packages/pyramid/config.py", line 491, in commit
self._ctx.execute_actions()
File "/home/viraptor/pyramid/lib/python2.6/site-packages/zope/configuration/config.py", line 626, in execute_actions
callable(*args, **kw)
File "/home/viraptor/pyramid/lib/python2.6/site-packages/pyramid/config.py", line 1291, in register
derived_view = deriver(view)
File "/home/viraptor/pyramid/lib/python2.6/site-packages/pyramid/config.py", line 2681, in __call__
self.mapped_view(view))))))))
File "/home/viraptor/pyramid/lib/python2.6/site-packages/pyramid/config.py", line 2624, in inner
wrapped_view = wrapped(self, view)
File "/home/viraptor/pyramid/lib/python2.6/site-packages/pyramid/config.py", line 2693, in mapped_view
mapped_view = mapper(**self.kw)(view)
File "/home/viraptor/pyramid/lib/python2.6/site-packages/pyramid/config.py", line 2860, in __call__
view = self.map_nonclass(view)
File "/home/viraptor/pyramid/lib/python2.6/site-packages/pyramid/config.py", line 2876, in map_nonclass
ronly = requestonly(view, self.attr)
File "/home/viraptor/pyramid/lib/python2.6/site-packages/pyramid/config.py", line 2966, in requestonly
if len(args) - len(defaults) == 1:
zope.configuration.config.ConfigurationExecutionError: <type 'exceptions.TypeError'>: object of type 'NoneType' has no len()
in:
('/home/viraptor/blah/blah/__init__.py', 22, 'main', "config.add_route('customer', '/customer/{customer_id}', view='blah.views.customer.view', view_renderer='customer_view.mak', view_permission='view', traverse='/customer/{customer_id}')")
这可能是什么原因?我最近甚至没有更改过该配置,只有应用程序的其余部分。
答案 0 :(得分:3)
我怀疑你遇到了金字塔新修订版中修复的错误;您的追溯表明args
或defaults
是None
,但除非args
不是None
,否则无法访问该代码分支,可能会defaults
1}}而不是None
。我发现以下对Pyramids的提交为defaults
添加了一个特定的测试:None:
https://github.com/Pylons/pyramid/commit/f168197609169fb01b65adeb3eb59d069000fe2c
我说你有一个没有任何默认值的方法,只有一个请求参数(method(self, request)
,解决办法是添加一个默认的关键字参数(method(self, request, dummy=None)
。
免责声明:还没有机会与Pyramid合作,所以我的分析完全基于Pyramid代码库。
答案 1 :(得分:0)
config.add_route
只接受1个位置参数,你的第二个参数应该与pattern
使用keyworded。
其次,我认为route
和traverse
的模式不一样。使用traverse
关键字,您可以确定root
应该从哪里开始。它在Configurator
API documentation中解释。
引发的错误异常可能会提供更多信息。