urllib

I’m playing around with the Stack Overflow API using Python. I’m trying to decode the gzipped responses that the API gives. import urllib, gzip url = urllib.urlopen('http://api.stackoverflow.com/1.0/badges/name') gzip.GzipFile(fileobj=url).read() According to the urllib2 documentation , urlopen “returns a file-like object”. However, when I run read() on the GzipFile object I’ve created using it, I get this error: AttributeError: addinfourl instance has no attribute 'tell' As far as I can tell, this is coming from the object returned by urlopen . It doesn’t appear to have seek either, as I get

I want to get response time when I use urllib . I made below code, but it is more than response time. Can I get the time using urllib or have any other method? import urllib import datetime def main(): urllist = [ "http://google.com", ] for url in urllist: opener = urllib.FancyURLopener({}) try: start = datetime.datetime.now() f = opener.open(url) end = datetime.datetime.now() diff = end - start print int(round(diff.microseconds / 1000)) except IOError, e: print 'error', url else: print f.getcode(), f.geturl() if __name__ == "__main__": main() Save yourself some hassle and use the requests

How can I urlencode a string with special chars æøå? ex. urllib.urlencode('http://www.test.com/q=testæøå') I get this error :(.. not a valid non-string sequence or mapping object You should pass dictionary to urlencode, not a string. See the correct example below: from urllib import urlencode print 'http://www.test.com/?' + urlencode({'q': 'testæøå'}) urlencode is intended to take a dictionary, for example: >>> q= u'\xe6\xf8\xe5' # u'æøå' >>> params= {'q': q.encode('utf-8')} >>> 'http://www.test.com/?'+urllib.urlencode(params) 'http://www.test.com/?q=%C3%A6%C3%B8%C3%A5' If you just want to URL

介绍 urllib.parse是为urllib包下面的一个模块，urllib的其它模块完全可以使用requests替代。但是urlli.parse我们是有必要了解的，因为该模块下面有很多操作url路径的方法 urlparse：拆分url from urllib import parse url = "https://www.baidu.com/s?wd=python" print(parse.urlparse(url)) # ParseResult(scheme='https', netloc='www.baidu.com', path='/s', params='', query='wd=python', fragment='') """ scheme:协议，比如http，https等等。 netloc:域名，这里是www.baidu.com path:路径，跟在域名后面 params:参数 query:查询条件 fragment:锚点，用于直接定位页面的下拉位置，跳转到网页的指定位置 """ scheme, netloc, path, params, query, fragment = parse.urlparse(url) print(f"协议:{scheme}") print(f"域名:{netloc}") print(f"路径:{path}") print(f"参数:

import urllib print urllib.urlopen('http://www.reefgeek.com/equipment/Controllers_&_Monitors/Neptune_Systems_AquaController/Apex_Controller_&_Accessories/').read() The above script works and returns the expected results while: import urllib2 print urllib2.urlopen('http://www.reefgeek.com/equipment/Controllers_&_Monitors/Neptune_Systems_AquaController/Apex_Controller_&_Accessories/').read() throws the following error: Traceback (most recent call last): File "<stdin>", line 1, in <module> File "/usr/lib/python2.5/urllib2.py", line 124, in urlopen return _opener.open(url, data) File "/usr/lib

I am trying to retrieve a 500mb file using Python, and I have a script which uses urllib.urlretrieve() . There seems to some network problem between me and the download site, as this call consistently hangs and fails to complete. However, using wget to retrieve the file tends to work without problems. What is the difference between urlretrieve() and wget that could cause this difference? Peter Lyons The answer is quite simple. Python's urllib and urllib2 are nowhere near as mature and robust as they could be. Even better than wget in my experience is cURL . I've written code that downloads