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社区首页 >问答首页 >从url上传文件到Flask Python 3服务器

从url上传文件到Flask Python 3服务器
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Stack Overflow用户
提问于 2020-09-03 02:32:24
回答 1查看 91关注 0票数 0

我想从URL下载一些文件到我的服务器上。例如,您有url并想要保存它。我只找到了从用户计算机上下载的示例,但它在我的情况下不起作用。我使用下面的代码在本地程序中下载文件

代码语言:javascript
复制
with urllib.request.urlopen(url) as response, open(file_name, 'wb') as out_file:
        shutil.copyfileobj(response, out_file)

但是当在服务器上运行时,它会产生这个错误:

代码语言:javascript
复制
Traceback (most recent call last):
  File "/usr/lib/python3.8/site-packages/flask/app.py", line 2446, in wsgi_app
    response = self.full_dispatch_request()
  File "/usr/lib/python3.8/site-packages/flask/app.py", line 1951, in full_dispatch_request
    rv = self.handle_user_exception(e)
  File "/usr/lib/python3.8/site-packages/flask/app.py", line 1820, in handle_user_exception
    reraise(exc_type, exc_value, tb)
  File "/usr/lib/python3.8/site-packages/flask/_compat.py", line 39, in reraise
    raise value
  File "/usr/lib/python3.8/site-packages/flask/app.py", line 1949, in full_dispatch_request
    rv = self.dispatch_request()
  File "/usr/lib/python3.8/site-packages/flask/app.py", line 1935, in dispatch_request
    return self.view_functions[rule.endpoint](**req.view_args)
  File "/home/DenShlk/mysite/flask_app.py", line 26, in hello_world
    save_pic_by_id(id)
  File "/home/DenShlk/mysite/flask_app.py", line 20, in save_pic_by_id
    with urllib.request.urlopen(pic_url) as response, open(file_name, 'wb') as out_file:
  File "/usr/lib/python3.8/urllib/request.py", line 222, in urlopen
    return opener.open(url, data, timeout)
  File "/usr/lib/python3.8/urllib/request.py", line 525, in open
    response = self._open(req, data)
  File "/usr/lib/python3.8/urllib/request.py", line 542, in _open
    result = self._call_chain(self.handle_open, protocol, protocol +
  File "/usr/lib/python3.8/urllib/request.py", line 502, in _call_chain
    result = func(*args)
  File "/usr/lib/python3.8/urllib/request.py", line 1360, in https_open
    return self.do_open(http.client.HTTPSConnection, req,
  File "/usr/lib/python3.8/urllib/request.py", line 1320, in do_open
    raise URLError(err)
urllib.error.URLError: <urlopen error Tunnel connection failed: 403 Forbidden>

谢谢你的帮助!

EN

回答 1

Stack Overflow用户

回答已采纳

发布于 2020-09-03 03:26:16

终于,我找到了真正的问题所在。我用的是pythonanywhere.com,有一个可以从免费账户访问的网站白名单。

票数 0
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页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接:

https://stackoverflow.com/questions/63711630

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