Python urllib2保持活着

问题描述

如何使用Python的urllib2创建保持活跃的HTTP请求？

解决方案

使用库。这包括支持HTTP 1.1和keepalive的urllib2的HTTP处理程序：

 >>> import urllib2 
>>>来自urlgrabber.keepalive导入HTTPHandler 
>>> keepalive_handler = HTTPHandler（）
>>> opener = urllib2.build_opener（keepalive_handler）
>>> urllib2.install_opener（opener）
>>> 
>>> fo = urllib2.urlopen（'http://www.python.org'）
  

注意：你应使用urlgrabber版本或更早版本，作为版本3.9.1中已删除模块

有一个针对Python 3的keepalive模块的。 / p>

How can I make a "keep alive" HTTP request using Python's urllib2?

解决方案

Use the urlgrabber library. This includes an HTTP handler for urllib2 that supports HTTP 1.1 and keepalive:

>>> import urllib2
>>> from urlgrabber.keepalive import HTTPHandler
>>> keepalive_handler = HTTPHandler()
>>> opener = urllib2.build_opener(keepalive_handler)
>>> urllib2.install_opener(opener)
>>>
>>> fo = urllib2.urlopen('http://www.python.org')

Note: you should use urlgrabber version 3.9.0 or earlier, as the keepalive module has been removed in version 3.9.1

There is a port of the keepalive module to Python 3.

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