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

Async现在是一个独立的模块:grequests。

请看这里:https://github.com/kennethreitz/grequests

还有:通过Python发送多个HTTP请求的理想方法?

安装:

$ pip install grequests

用法:

建立一个堆栈:

import grequests

urls = [
    'http://www.heroku.com',
    'http://tablib.org',
    'http://httpbin.org',
    'http://python-requests.org',
    'http://kennethreitz.com'
]

rs = (grequests.get(u) for u in urls)

发送堆栈

grequests.map(rs)

结果如下所示

[<Response [200]>, <Response [200]>, <Response [200]>, <Response [200]>, <Response [200]>]

grequest似乎没有设置并发请求的限制,即当多个请求被发送到同一个服务器时。

其他回答

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)

from threading import Thread

threads=list()

for requestURI in requests:
    t = Thread(target=self.openURL, args=(requestURI,))
    t.start()
    threads.append(t)

for thread in threads:
    thread.join()

...

def openURL(self, requestURI):
    o = urllib2.urlopen(requestURI, timeout = 600)
    o...

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。

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

Async现在是一个独立的模块:grequests。

请看这里:https://github.com/kennethreitz/grequests

还有:通过Python发送多个HTTP请求的理想方法?

安装:

$ pip install grequests

用法:

建立一个堆栈:

import grequests

urls = [
    'http://www.heroku.com',
    'http://tablib.org',
    'http://httpbin.org',
    'http://python-requests.org',
    'http://kennethreitz.com'
]

rs = (grequests.get(u) for u in urls)

发送堆栈

grequests.map(rs)

结果如下所示

[<Response [200]>, <Response [200]>, <Response [200]>, <Response [200]>, <Response [200]>]

grequest似乎没有设置并发请求的限制,即当多个请求被发送到同一个服务器时。