这会调用请求函数

Question

我正在编写一个脚本列表并解析信息。

它适用于大多数网站，但它有些令人窒息 “UnicodeEncodeError：'ascii'编解码器无法编码位置13中的字符'\ xe9'：序数不在范围内（128）”

它在client.py上停止，它是python3上urlib的一部分

确切的链接是： http://finance.yahoo.com/news /咖啡厅-成长快于快餐-同行-144512056.html

这里有很多类似的贴子，但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

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

追溯是：

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)

任何帮助都赞赏它让我疯狂，我想我已经尝试过x.decode和类似的所有组合

（如果可能，我可以忽略有问题的人物。）

Answer 1

使用percent-encoded URL：

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

我通过将浏览器指向

找到了上述百分比编码的URL

http://finance.yahoo.com/news/cafés-growing-faster-than-fast-food-peers-144512056.html

转到页面，然后复制并粘贴由浏览器提供的编码网址返回文本编辑器。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：

from urllib import parse

link = 'http://finance.yahoo.com/news/cafés-growing-faster-than-fast-food-peers-144512056.html'

scheme, netloc, path, query, fragment = parse.urlsplit(link)
path = parse.quote(path)
link = parse.urlunsplit((scheme, netloc, path, query, fragment))

产生

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

Answer 2

您的网址包含无法用ASCII字符表示的字符。

您必须确保所有字符都经过正确的网址编码;以urllib.parse.quote_plus为例;它将使用UTF-8 URL编码的转义来表示任何非ASCII字符。

UnicodeEncodeError：＆＃39; ascii＆＃39;编解码器不能对字符＆＃39; \ xe9＆＃39;进行编码。 - 使用urlib.request python3时

这会调用请求函数

2 个答案: