urllib request.urlretrieve 错误 400

Question

我正在尝试从 url

下载图片

request.urlretrieve('https://website.com/image.jpg', "test.jpg")

以上有效。但是，当我尝试使用时：

a = 'https://website.com/image.jpg'
request.urlretrieve(a, "test.jpg")

它给我错误 400。如何在不导致错误的情况下将 URL 放入其中？

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

400 Bad Request 错误的一个常见原因是 URL 输入错误。尝试 print(a) 检查它是否确实包含您的想法。

以下几行对我有用 (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")

您也可以尝试使用另一个 URL，例如上面的那个，检查它是否特定于您的 URL（这可能与 DNS 问题或服务器问题有关）

urllib request.urlretrieve 错误 400

urllib request.urlretrieve error 400

urllib