用于 Sinatra 风格 URL 路由的 Python 等效 Ruby 块
有没有办法在 Python 中重新创建 Sinatra 的 URL 路由？有什么理由可以解释为什么这可能是不可取的吗？
来自 Sinatra：
get '/' do
'Hello world!'
end
来自 Flask（使用装饰器进行路由）：
@app.route("/")
def hello():
return "Hello World!"
Sinatra 通过 Ruby 块实现了这种简洁性：
def get(path, opts={}, &block)
conditions = @conditions.dup
route('GET', path, opts, &block)
@conditions = conditions
route('HEAD', path, opts, &block)
end
我认为 Python 没有与 Ruby 块完全相同的功能，但有一些方法可以重新创建功能。这怎么办？
Is there a way of recreating Sinatra's URL routing in Python? And are there any reasons why this might not be desirable?
From Sinatra:
get '/' do
'Hello world!'
end
From Flask (using decorators for routing):
@app.route("/")
def hello():
return "Hello World!"
Sinatra achieves this succinctness through Ruby blocks:
def get(path, opts={}, &block)
conditions = @conditions.dup
route('GET', path, opts, &block)
@conditions = conditions
route('HEAD', path, opts, &block)
end
I gather that Python does not have an exact equivalent of Ruby blocks, but that there are ways of recreating the functionality. How might this be done?
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正如你所说，Python 没有像 ruby 块这样的东西。装饰器是常用的路由解决方案。另一种方法是创建一个包含路由的列表/字典，但由于您似乎希望路由定义位于底层代码旁边，因此您需要使用装饰器方法。
As you said, Python doesn't have something like ruby blocks. Decorators are the commonly used solution for routing. The other way would be creating a list/dict containing the routes but since you seem to want the route definitions next to the underlying code, the decorator way is what you'll want to use.
另一种方法是使用元类，就像在 webpy 的 web.autoapplication 中完成的那样，源代码。
Another way would be using metaclasses, as it is done in webpy's web.autoapplication, source code for that.
那么，对于你的第二个问题“是否有任何理由表明这可能是不可取的？”。
url_for(some_function)的操作，从而可以轻松地重构网站。So, to your second question of 'And are there any reasons why this might not be desirable?'.
url_for(some_function), which makes it easy to restructure the site.