请求会在几秒钟后执行重试方法吗？

Question

这是我的代码.

import requests,time
proxies = {'http':'36.33.1.177:21219'}
url='http://218.94.78.61:8080/newPub/service/json/call?serviceName=sysBasicManage&methodName=queryOutputOtherPollutionList&paramsJson=%7B%22ticket%22:%22451a9846-058b-4944-86c6-fccafdb7d8d0%22,%22parameter%22:%7B%22monitorSiteType%22:%2202%22,%22enterpriseCode%22:%22320100000151%22,%22monitoringType%22:%222%22%7D%7D'

i = 0
a = requests.adapters.HTTPAdapter(max_retries=10)
s = requests.Session()
s.mount(url, a)
for x in xrange(1,1000):
    time.sleep(1)
    print x
    try:
        r= s.get(url,proxies=proxies)
        print r
    except Exception as ee:
        i = i + 1
        print ee
        print 'i=%s' % i

代理有点不稳定,所以我设置了max_retries,但它有时候仍然有异常,所以在每次重试后有一些secondes执行的方法吗？

Answer 1

只是requests图书馆,这是不可能的.但是,您可以使用外部库,如退避.

backoff提供一个装饰器,你将它包装在你的功能周围.示例代码:

@backoff.on_exception(backoff.constant,
                      requests.exceptions.RequestException,
                      max_tries=10, interval=10)
def get_url(url):
    return requests.get(url)

上面的代码等待10秒,以便在每次例外时进行下一次重试,requests.exceptions.RequestException并尝试10次,如中所述max_tries.