UnicodeEncodeError: 'ascii' codec can't encode character '\xe9' - -when using urlib.request python3

Question

我正在编写一个脚本，该脚本转到链接列表并解析信息。

它适用于大多数站点，但它在某些站点上令人窒息，“UnicodeEncodeError：'ascii' 编解码器无法在位置 13 中编码字符 '\xe9'：序号不在范围内（128）”

它在 client.py 上停止，它是 python3 上 urlib 的一部分

确切链接是：http : //finance.yahoo.com/news/cafés-growth-faster-than-fast-food-peers-144512056.html

这里有很多类似的帖子，但似乎没有一个答案对我有用。

我的代码是：

from urllib import request

def __request(link,debug=0):      

try:
    html = request.urlopen(link, timeout=35).read() #made this long as I was getting lots of timeouts
    unicode_html = html.decode('utf-8','ignore')

# NOTE the except HTTPError must come first, otherwise except URLError will also catch an HTTPError.
except HTTPError as e:
    if debug:
        print('The server couldn\'t fulfill the request for ' + link)
        print('Error code: ', e.code)
    return ''
except URLError as e:
    if isinstance(e.reason, socket.timeout):
        print('timeout')
        return ''    
else:
    return unicode_html

这会调用请求函数

链接 = ' http://finance.yahoo.com/news /cafés-gromming-faster-than-fast-food-peers-144512056.html' 页面 = __request(link)

回溯是：

Traceback (most recent call last):
  File "<string>", line 250, in run_nodebug
  File "C:\reader\get_news.py", line 276, in <module>
    main()
  File "C:\reader\get_news.py", line 255, in main
    body = get_article_body(item['link'],debug=0)
  File "C:\reader\get_news.py", line 155, in get_article_body
    page = __request('na',url)
  File "C:\reader\get_news.py", line 50, in __request
    html = request.urlopen(link, timeout=35).read()
  File "C:\Python33\Lib\urllib\request.py", line 156, in urlopen
    return opener.open(url, data, timeout)
  File "C:\Python33\Lib\urllib\request.py", line 469, in open
    response = self._open(req, data)
  File "C:\Python33\Lib\urllib\request.py", line 487, in _open
    '_open', req)
  File "C:\Python33\Lib\urllib\request.py", line 447, in _call_chain
    result = func(*args)
  File "C:\Python33\Lib\urllib\request.py", line 1268, in http_open
    return self.do_open(http.client.HTTPConnection, req)
  File "C:\Python33\Lib\urllib\request.py", line 1248, in do_open
    h.request(req.get_method(), req.selector, req.data, headers)
  File "C:\Python33\Lib\http\client.py", line 1061, in request
    self._send_request(method, url, body, headers)
  File "C:\Python33\Lib\http\client.py", line 1089, in _send_request
    self.putrequest(method, url, **skips)
  File "C:\Python33\Lib\http\client.py", line 953, in putrequest
    self._output(request.encode('ascii'))
UnicodeEncodeError: 'ascii' codec can't encode character '\xe9' in position 13: ordinal not in range(128)

任何帮助表示赞赏这让我发疯，我想我已经尝试了 x.decode 和类似的所有组合

（如果可能的话，我可以忽略有问题的字符。）

Answer 1

使用百分比编码的 URL：

\n\n

link = \'http://finance.yahoo.com/news/caf%C3%A9s-growing-faster-than-fast-food-peers-144512056.html\'\n

\n\n\n\n

我通过将浏览器指向上面的百分比编码 URL

\n\n

http://finance.yahoo.com/news/caf\xc3\xa9s-growing-faster-than-fast-food-peers-144512056.html\n

\n\n

转到该页面，然后将浏览器提供的编码 URL 复制并粘贴回文本编辑器中。但是，您可以使用以下命令以编程方式生成百分比编码的 URL：

\n\n

from urllib import parse\n\nlink = \'http://finance.yahoo.com/news/caf\xc3\xa9s-growing-faster-than-fast-food-peers-144512056.html\'\n\nscheme, netloc, path, query, fragment = parse.urlsplit(link)\npath = parse.quote(path)\nlink = parse.urlunsplit((scheme, netloc, path, query, fragment))\n

\n\n

这产生

\n\n

http://finance.yahoo.com/news/caf%C3%A9s-growing-faster-than-fast-food-peers-144512056.html\n

\n