Return (new, timeout=req.timeout)įile "/usr/local/lib/python3.4/urllib/request.py", line 455, in openįile "/usr/local/lib/python3.4/urllib/request.py", line 478, in _openįile "/usr/local/lib/python3.4/urllib/request. 'http', request, response, code, msg, hdrs)įile "/usr/local/lib/python3.4/urllib/request.py", line 493, in errorįile "/usr/local/lib/python3.4/urllib/request.py", line 433, in _call_chainįile "/usr/local/lib/python3.4/urllib/request.py", line 676, in http_error_302 Note: All code in this answer was created for Python 3 see end of answer to use this code with Python 2. With contextlib.closing(urlopen(url, data)) as fp:įile "/usr/local/lib/python3.4/urllib/request.py", line 153, in urlopenįile "/usr/local/lib/python3.4/urllib/request.py", line 461, in openįile "/usr/local/lib/python3.4/urllib/request.py", line 571, in http_response Python 3 A Simple, Customizable Progress Bar Heres an aggregate of many of the answers below that I use regularly (no imports required). Download Python 3.9.1 for Windows - Windows Desktop Programming Languages Python for Windows Python 3.9.1 for Windows Python for Windows 3.9.1 Python User rating Installed through our safe & fast downloader ( more info) Download Technical Title: Python 3.9. My error:- Traceback (most recent call last):įile "/usr/local/lib/python3.4/urllib/request.py", line 178, in urlretrieve
Since it is now forwarded to https with 302 redirection. My script import the following library and uses the urlretrive to get the file previously. I have a script on python3.4 and it has been fine until the website I download the file from decides to use https and now I am getting error but can't figure out how I can retrive the file.
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