Pyramid Traversal __name__匹配视图名称

Question

在Traversal金字塔应用程序中,如何处理__name__与视图名称匹配的资源？

如果我想进入资源的视图可调用"视图",我将使用如下的URL路径:/foo/bar/view.它遍历resource_tree,如下所示:

RootFactory(request) => RootFactory
RootFactory['foo']   => Foo
Foo['bar']           => Bar
Bar['view']          => KeyError

...并且因为它无法遍历Bar&'view',所以它假定'view'是视图名称,并匹配我的视图可调用

@view_config(name='view')
def view(context, request):
    return Response(context.__name__)

获取我使用的路径的URL request.resource_url(resource, "view".

但是,如果我有这样的资源Bar.__name__ = "view",如何在Foo上解析"查看"的URL？

# path: /foo/view
RootFactory(request) => RootFactory
RootFactory['foo']   => Foo  # ideally stop here with view_name="view"
Foo['view']          => Bar.__name__ = "view"
# all parts of path matched so view_name="" and context=Bar

理想情况下,在这种情况下,/foo/view会指向view(Foo, request),并/foo/view/view指出在view(Bar, request)哪里Bar.__name__ == "view".

有没有办法处理这个没有写入检测__name__视图名称之间的冲突？

Answer 1

我最终找到的解决方案是添加任意层容器资源。

以下是资源的示例结构：

class Foo(object):

    def __init__(self, parent, name):
        self.__parent__ = parent
        self.__name__ = name

    def __getitem__(self, key):
        if key == "bar":
            return BarContainer(self, "bar")
        else:
            raise KeyError()

class BarContainer(object):

    def __init__(self, parent, name):
        self.__parent__ = parent
        self.__name__ = name

    def __getitem__(self, key):
        return Bar(self, key)

class Bar(object):

    def __init__(self, parent, name):
        self.__parent__ = parent
        self.__name__ = name

该问题提到想要生成可调用标题视图的资源 URL。这样BarContainer只需添加一个额外的斜杠：

/{foo}/bar/{bar}/<view_name>    
/a/view          => view(context=Foo(name='a'), request)
/a/bar/view      => view(context=BarContainer(name='bar'), request)
/a/bar/b/view    => view(context=Bar(name='b'), request)