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How can I make a "keep alive" HTTP request using Python's urllib2?
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urllib2 is deprecated in python 3. x. use urllib instaed.
1) urllib2 can accept a Request object to set the headers for a URL request, urllib accepts only a URL. 2) urllib provides the urlencode method which is used for the generation of GET query strings, urllib2 doesn't have such a function. This is one of the reasons why urllib is often used along with urllib2.
Simple urllib2 script urlopen('http://python.org/') print "Response:", response # Get the URL. This gets the real URL. print "The URL is: ", response. geturl() # Getting the code print "This gets the code: ", response.
NOTE: urllib2 is no longer available in Python 3 You can get more idea about urllib.
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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