Python 3.4中的"异步"

Question

aiohttp的入门文档提供了以下客户端示例:

import asyncio
import aiohttp

async def fetch_page(session, url):
    with aiohttp.Timeout(10):
        async with session.get(url) as response:
            assert response.status == 200
            return await response.read()

loop = asyncio.get_event_loop()
with aiohttp.ClientSession(loop=loop) as session:
    content = loop.run_until_complete(
        fetch_page(session, 'http://python.org'))
    print(content)

他们为Python 3.4用户提供以下注释:

如果您使用的是Python 3.4,请使用@coroutine装饰器替换a yield和async def.

如果我遵循这些说明,我会得到:

import aiohttp
import asyncio

@asyncio.coroutine
def fetch(session, url):
    with aiohttp.Timeout(10):
        async with session.get(url) as response:
            return (yield from response.text())

if __name__ == '__main__':
    loop = asyncio.get_event_loop()
    with aiohttp.ClientSession(loop=loop) as session:
        html = loop.run_until_complete(
            fetch(session, 'http://python.org'))
        print(html)

但是,这不会运行,因为async withPython 3.4不支持:

$ python3 client.py 
  File "client.py", line 7
    async with session.get(url) as response:
             ^
SyntaxError: invalid syntax

如何将async with语句翻译为与Python 3.4一起使用？

Answer 1

只是不要将结果session.get()用作上下文管理器; 直接用它作为协程.生成的请求上下文管理器session.get()通常会在退出时释放请求,但使用response.text()时也是如此,因此您可以忽略此处:

@asyncio.coroutine
def fetch(session, url):
    with aiohttp.Timeout(10):
        response = yield from session.get(url)
        return (yield from response.text())

这里返回的请求包装器没有必需的异步方法(__aenter__和__aexit__),当不使用Python 3.5时它们完全省略(参见相关的源代码).

如果你在session.get()呼叫和访问response.text()等待之间有更多的声明,你可能想要使用一个try:..finally:来释放连接; 如果发生异常,Python 3.5发行版上下文管理器也会关闭响应.因为yield from response.release()这里需要a ,所以在Python 3.4之前不能将它封装在上下文管理器中:

import sys

@asyncio.coroutine
def fetch(session, url):
    with aiohttp.Timeout(10):
        response = yield from session.get(url)
        try:
            # other statements
            return (yield from response.text())
        finally:
            if sys.exc_info()[0] is not None:
                # on exceptions, close the connection altogether
                response.close()
            else:
                yield from response.release()

Answer 2

aiohttp使用3.4语法实现的示例.基于json客户端示例,您的函数将是:

@asyncio.coroutine
def fetch(session, url):
    with aiohttp.Timeout(10):
        resp = yield from session.get(url)
        try:
            return (yield from resp.text())
        finally:
            yield from resp.release()

UPD:

请注意,Martijn的解决方案适用于简单的情况,但在特定情况下可能会导致不必要的行为:

@asyncio.coroutine
def fetch(session, url):
    with aiohttp.Timeout(5):
        response = yield from session.get(url)

        # Any actions that may lead to error:
        1/0

        return (yield from response.text())

# exception + warning "Unclosed response"

除了例外,你还会收到警告"Unclosed response".这可能会导致复杂应用程序中的连接泄漏.如果您手动调用resp.release()/,您将避免此问题resp.close():

@asyncio.coroutine
def fetch(session, url):
    with aiohttp.Timeout(5):
        resp = yield from session.get(url)
        try:

            # Any actions that may lead to error:
            1/0

            return (yield from resp.text())
        except Exception as e:
            # .close() on exception.
            resp.close()
            raise e
        finally:
            # .release() otherwise to return connection into free connection pool.
            # It's ok to release closed response:
            # https://github.com/KeepSafe/aiohttp/blob/master/aiohttp/client_reqrep.py#L664
            yield from resp.release()

# exception only

我认为最好遵循官方示例(和__aexit__ 实现)并明确调用resp.release()/ resp.close().