python - 如何使用 asyncio 添加连接超时？

Question

我想快速连接到许多不同站点的列表。我使用 asyncio 以异步方式执行此操作，现在想如果响应时间太长，则应在何时忽略连接时添加超时。

我该如何实现？

import ssl
import asyncio
from contextlib import suppress
from concurrent.futures import ThreadPoolExecutor
import time


@asyncio.coroutine
def run():
    while True:
        host = yield from q.get()
        if not host:
            break

        with suppress(ssl.CertificateError):
            reader, writer = yield from asyncio.open_connection(host[1], 443, ssl=True) #timout option?
            reader.close()
            writer.close()


@asyncio.coroutine
def load_q():
    # only 3 entries for debugging reasons
    for host in [[1, 'python.org'], [2, 'qq.com'], [3, 'google.com']]:
        yield from q.put(host)
    for _ in range(NUM):
        q.put(None)


if __name__ == "__main__":
    NUM = 1000
    q = asyncio.Queue()

    loop = asyncio.get_event_loop()
    loop.set_default_executor(ThreadPoolExecutor(NUM))

    start = time.time()
    coros = [asyncio.async(run()) for i in range(NUM)]
    loop.run_until_complete(load_q())
    loop.run_until_complete(asyncio.wait(coros))
    end = time.time()
    print(end-start)

(旁注:有人知道如何优化它吗？)

Answer 1

您可以将对open_connection 的调用包装在asyncio.wait_for 中，它允许您指定超时:

    with suppress(ssl.CertificateError):
        fut = asyncio.open_connection(host[1], 443, ssl=True)
        try:
            # Wait for 3 seconds, then raise TimeoutError
            reader, writer = yield from asyncio.wait_for(fut, timeout=3)
        except asyncio.TimeoutError:
            print("Timeout, skipping {}".format(host[1]))
            continue

请注意，当引发 TimeoutError 时，open_connection 协同程序也会被取消。如果您不希望它被取消(尽管我认为您确实在这种情况下希望它被取消)，您已将调用包装在 asyncio.shield 中。 .