我尝试了python请求库文档中提供的示例。

使用async.map(rs),我获得了响应代码,但我想获得所请求的每个页面的内容。例如,这是行不通的:

out = async.map(rs)
print out[0].content

当前回答

上面的答案都没有帮助我,因为他们假设你有一个预定义的请求列表,而在我的情况下,我需要能够侦听请求和异步响应(类似于它在nodejs中的工作方式)。

def handle_finished_request(r, **kwargs):
    print(r)


# while True:
def main():
    while True:
        address = listen_to_new_msg()  # based on your server

        # schedule async requests and run 'handle_finished_request' on response
        req = grequests.get(address, timeout=1, hooks=dict(response=handle_finished_request))
        job = grequests.send(req)  # does not block! for more info see https://stackoverflow.com/a/16016635/10577976


main()

handle_finished_request回调函数将在收到响应时被调用。注意:由于某些原因,超时(或无响应)在这里不会触发错误

这个简单的循环可以触发异步请求,类似于它在nodejs服务器中的工作方式

其他回答

上面的答案都没有帮助我,因为他们假设你有一个预定义的请求列表,而在我的情况下,我需要能够侦听请求和异步响应(类似于它在nodejs中的工作方式)。

def handle_finished_request(r, **kwargs):
    print(r)


# while True:
def main():
    while True:
        address = listen_to_new_msg()  # based on your server

        # schedule async requests and run 'handle_finished_request' on response
        req = grequests.get(address, timeout=1, hooks=dict(response=handle_finished_request))
        job = grequests.send(req)  # does not block! for more info see https://stackoverflow.com/a/16016635/10577976


main()

handle_finished_request回调函数将在收到响应时被调用。注意:由于某些原因,超时(或无响应)在这里不会触发错误

这个简单的循环可以触发异步请求,类似于它在nodejs服务器中的工作方式

我测试了两个请求——未来请求和请求请求。Grequests速度更快,但会带来猴子补丁和依赖关系的其他问题。请求-期货比请求慢几倍。我决定编写自己的请求,并简单地将请求包装到ThreadPoolExecutor中,它几乎和grequest一样快,但没有外部依赖。

import requests
import concurrent.futures

def get_urls():
    return ["url1","url2"]

def load_url(url, timeout):
    return requests.get(url, timeout = timeout)

with concurrent.futures.ThreadPoolExecutor(max_workers=20) as executor:

    future_to_url = {executor.submit(load_url, url, 10): url for url in     get_urls()}
    for future in concurrent.futures.as_completed(future_to_url):
        url = future_to_url[future]
        try:
            data = future.result()
        except Exception as exc:
            resp_err = resp_err + 1
        else:
            resp_ok = resp_ok + 1

我已经使用python请求异步调用github的gist API有一段时间了。

举个例子,请看下面的代码:

https://github.com/davidthewatson/flasgist/blob/master/views.py#L60-72

这种风格的python可能不是最清晰的例子,但我可以向您保证代码是有效的。如果这让你感到困惑,请告诉我,我会记录下来。

也许请求-期货是另一种选择。

from requests_futures.sessions import FuturesSession

session = FuturesSession()
# first request is started in background
future_one = session.get('http://httpbin.org/get')
# second requests is started immediately
future_two = session.get('http://httpbin.org/get?foo=bar')
# wait for the first request to complete, if it hasn't already
response_one = future_one.result()
print('response one status: {0}'.format(response_one.status_code))
print(response_one.content)
# wait for the second request to complete, if it hasn't already
response_two = future_two.result()
print('response two status: {0}'.format(response_two.status_code))
print(response_two.content)

办公文档中也有建议。如果你不想卷入gevent,这是一个不错的选择。

你可以使用httpx。

import httpx

async def get_async(url):
    async with httpx.AsyncClient() as client:
        return await client.get(url)

urls = ["http://google.com", "http://wikipedia.org"]

# Note that you need an async context to use `await`.
await asyncio.gather(*map(get_async, urls))

如果你想要一个函数式语法,gamla库将其包装到get_async中。

然后你就可以


await gamla.map(gamla.get_async(10))(["http://google.com", "http://wikipedia.org"])

10是超时时间,单位是秒。

(声明:我是作者)