在 Pyramid 中调用另一个视图

问题描述

我的目标:在 Pyramid 中，调用另一个视图可调用对象，并在不知道该视图可调用对象的任何详细信息的情况下返回一个 Response 对象.

My goal: In Pyramid, to call another view-callable, and to get a Response object back without knowing any details about that view-callable.

在我的 Pyramid 应用程序中，假设我有一个使用 view_config 装饰器定义的视图foo":

In my Pyramid application, say I have a view "foo" which is defined using a view_config decorator:

@view_config(route_name="foo",
             renderer="foo.jinja2")
def foo_view(request):
    return {"whereami" : "foo!"}

现在说我想将bar"路由到暂时执行相同操作的视图，因此它在内部调用 foo_view 并返回其响应:

Now say that I want to route "bar" to a view that does the same thing for the time being, so it internally calls foo_view and returns its Response:

@view_config(route_name="bar")
def bar_view(request):
   return foo_view(request)

...但是等等！这是行不通的，因为 foo_view 不返回 Response，它的 renderer 会返回.

...but wait! That doesn't work, since foo_view doesn't return a Response, its renderer does.

所以，这会起作用:

@view_config(route_name="bar",
             renderer="foo.jinja2")
def bar_view(request):
    return foo_view(request)

因为它将应用与 foo_view 相同的渲染器.但这很糟糕，因为我现在必须通过复制渲染器值来重复自己，并且必须知道被调用视图的渲染器.

as it will apply the same renderer as foo_view did. But this is bad, as I now must repeat myself by copying the renderer value AND having to know the renderer of the view being called.

所以，我希望 Pyramid 中有一些可用的函数，它允许调用另一个可调用的视图并返回一个 Response 对象，而无需知道或关心它是如何呈现的:

So, I am going to hope that there is some function available in Pyramid that allows calling another view-callable and getting a Response object back without knowing or caring how it was rendered:

@view_config(route_name="bar")
def bar_view(request):
    response = some_function_that_renders_a_view_callable(foo_view, request)
    return response

some_function_that_renders_a_view_callable 是什么?

pyramid.views.render_view 似乎按名称搜索视图；我不想给我的观点命名.

pyramid.views.render_view appears to search for a view by name; I don't want to give my views names.

(注意:返回 HTTPFound 导致客户端重定向到目标路由是我试图避免的.我想内部"重定向).

(Note: Returning HTTPFound to cause the client to redirect to the target route is what I am trying avoid. I want to "internally" redirect).

推荐答案

是的.有一些顾虑

• 不返回响应
• 谓词/渲染器
• 权限
• 与旧请求关联的请求属性

这就是为什么你不应该从视图中调用视图作为函数，除非你知道你在做什么

Thats why you should not call view from view as function, unless you know what you doing

金字塔创建者为服务器端重定向做了很棒的工具 - http://docs.pylonsproject.org/projects/pyramid/en/latest/narr/subrequest.html

Pyramid creators did awesome tool for server side redirect - http://docs.pylonsproject.org/projects/pyramid/en/latest/narr/subrequest.html

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