失败时如何重试urllib2.request? [英] How to retry urllib2.request when fails?
问题描述
当 urllib2.request
达到超时时,会引发 urllib2.URLError
异常。
重试建立连接的pythonic方法是什么?
When urllib2.request
reaches timeout, a urllib2.URLError
exception is raised.
What is the pythonic way to retry establishing a connection?
推荐答案
我会使用 retry 装饰器。还有其他一些,但是这个很好用。使用方法如下:
I would use a retry decorator. There are other ones out there, but this one works pretty well. Here's how you can use it:
@retry(urllib2.URLError, tries=4, delay=3, backoff=2)
def urlopen_with_retry():
return urllib2.urlopen("http://example.com")
如果引发 URLError
,它将重试该函数。检查上面的链接以获取有关参数的文档,但基本上,它最多可重试4次,每次的指数退避延迟都加倍,例如3秒,6秒,12秒。
This will retry the function if URLError
is raised. Check the link above for documentation on the parameters, but basically it will retry a maximum of 4 times, with an exponential backoff delay doubling each time, e.g. 3 seconds, 6 seconds, 12 seconds.
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