urllib request.urlretrieve error 400 - urllib

Question

I'm trying to download an image from a url
request.urlretrieve('https://website.com/image.jpg', "test.jpg")
The above works. However when I try to use:
a = 'https://website.com/image.jpg'
request.urlretrieve(a, "test.jpg")
It gives me error 400. How do I get the URL in there without causing an error?
File "/usr/lib/python3.5/urllib/request.py", line 188, in urlretrieve
with contextlib.closing(urlopen(url, data)) as fp:
File "/usr/lib/python3.5/urllib/request.py", line 163, in urlopen
return opener.open(url, data, timeout)
File "/usr/lib/python3.5/urllib/request.py", line 472, in open
response = meth(req, response)
File "/usr/lib/python3.5/urllib/request.py", line 582, in http_response
'http', request, response, code, msg, hdrs)
File "/usr/lib/python3.5/urllib/request.py", line 510, in error
return self._call_chain(*args)
File "/usr/lib/python3.5/urllib/request.py", line 444, in _call_chain
result = func(*args)
File "/usr/lib/python3.5/urllib/request.py", line 590, in http_error_default
raise HTTPError(req.full_url, code, msg, hdrs, fp)
urllib.error.HTTPError: HTTP Error 400: Bad Request

Answer 1

A common reason for a 400 Bad Request error is because the URL was typed wrong. Try to print(a) to check if it indeed contains what you think.
the following lines work for me (python 3.5):
import urllib.request
# without variable
urllib.request.urlretrieve("https://www.google.fr/images/branding/googlelogo/2x/googlelogo_color_272x92dp.png", "test.png")
# with variable
google_pic_url = "https://www.google.fr/images/branding/googlelogo/2x/googlelogo_color_272x92dp.png"
urllib.request.urlretrieve(google_pic_url, "test2.png")
You can also try with another URL, like the one above to check if it's specific with your URL (this could be related to DNS issues, or server issues)