python - aiohttp:装饰器序列链

Question

考虑以下代码:

from aiohttp_mako import template

def authorize():
    def wrapper(func):
        @asyncio.coroutine
        @functools.wraps(func)
        def wrapped(*args):
            allowed = # Some auth stuff
            if not allowed:
                return HTTPUnauthorized()
            return func()
        return wrapped
    return wrapper


@authorize()
@template('admin.mak')
async def admin(request):
    return dict(ok=True)

我希望 authorize() ' wrapper得到template装饰器作为其 func这样我就可以返回 Response它在我的 authorize 中生成装饰器。但是authorize() ' wrapper需要 admin()协程作为 func最后是

File "/Users/etemin/virtualenvs/onlinelux/lib/python3.5/site-packages/aiohttp/web.py", line 306, in _handle
    resp = yield from handler(request)
File "/Users/etemin/virtualenvs/onlinelux/lib/python3.5/site-packages/aiohttp_session/__init__.py", line 134, in middleware
    raise RuntimeError("Expect response, not {!r}", type(response))
RuntimeError: ('Expect response, not {!r}', <class 'generator'>)

因为它试图返回协程。我应该如何返回 template装饰器？

Answer 1

您已经包装了一个协同例程，因此您需要等待该协同例程(从中产生):

def authorize():
    def wrapper(func):
        @asyncio.coroutine
        @functools.wraps(func)
        def wrapped(*args):
            allowed = # Some auth stuff
            if not allowed:
                return HTTPUnauthorized()
            return yield from func()
        return wrapped
    return wrapper

因为您已经在使用 async/await 语法，所以我在这里也只使用它而不使用 @asyncio.coroutine :

def authorize():
    async def wrapper(func):
        @functools.wraps(func)
        async def wrapped(*args):
            allowed = # Some auth stuff
            if not allowed:
                return HTTPUnauthorized()
            return await func()
        return wrapped
    return wrapper

请注意，我在那里等待 func()，并返回了结果。