Question

我是python的新手，正在使用python 3.5.0。我试图按如下方式实现一个简单的代码：

import urllib.request
page = urllib.request.urlopen("http://www.google.com",timeout=20)
text = page.read().decode("utf8")
print(text)

但出乎意料的是我收到了以下错误：

Traceback (most recent call last):
  File "C:\Users\Dell\AppData\Local\Programs\Python\Python35-32\lib\urllib\request.py", line 1240, in do_open
    h.request(req.get_method(), req.selector, req.data, headers)
  File "C:\Users\Dell\AppData\Local\Programs\Python\Python35-32\lib\http\client.py", line 1083, in request
    self._send_request(method, url, body, headers)
  File "C:\Users\Dell\AppData\Local\Programs\Python\Python35-32\lib\http\client.py", line 1128, in _send_request
    self.endheaders(body)
  File "C:\Users\Dell\AppData\Local\Programs\Python\Python35-32\lib\http\client.py", line 1079, in endheaders
    self._send_output(message_body)
  File "C:\Users\Dell\AppData\Local\Programs\Python\Python35-32\lib\http\client.py", line 911, in _send_output
    self.send(msg)
  File "C:\Users\Dell\AppData\Local\Programs\Python\Python35-32\lib\http\client.py", line 854, in send
    self.connect()
  File "C:\Users\Dell\AppData\Local\Programs\Python\Python35-32\lib\http\client.py", line 826, in connect
    (self.host,self.port), self.timeout, self.source_address)
  File "C:\Users\Dell\AppData\Local\Programs\Python\Python35-32\lib\socket.py", line 707, in create_connection
    raise err
  File "C:\Users\Dell\AppData\Local\Programs\Python\Python35-32\lib\socket.py", line 698, in create_connection
    sock.connect(sa)
socket.timeout: timed out

During handling of the above exception, another exception occurred:

Traceback (most recent call last):
  File "D:\Python\Test_Run.py", line 2, in <module>
    page = urllib.request.urlopen("http://www.google.com",timeout=20)
  File "C:\Users\Dell\AppData\Local\Programs\Python\Python35-32\lib\urllib\request.py", line 162, in urlopen
    return opener.open(url, data, timeout)
  File "C:\Users\Dell\AppData\Local\Programs\Python\Python35-32\lib\urllib\request.py", line 465, in open
    response = self._open(req, data)
  File "C:\Users\Dell\AppData\Local\Programs\Python\Python35-32\lib\urllib\request.py", line 483, in _open
    '_open', req)
  File "C:\Users\Dell\AppData\Local\Programs\Python\Python35-32\lib\urllib\request.py", line 443, in _call_chain
    result = func(*args)
  File "C:\Users\Dell\AppData\Local\Programs\Python\Python35-32\lib\urllib\request.py", line 1268, in http_open
    return self.do_open(http.client.HTTPConnection, req)
  File "C:\Users\Dell\AppData\Local\Programs\Python\Python35-32\lib\urllib\request.py", line 1242, in do_open
    raise URLError(err)
urllib.error.URLError: <urlopen error timed out>

我已连接到我的大学网络。这就是我得到这个的原因吗？我该怎么办？

Answer 1

MAX_RETRY = 5

def get_html(html_url, timeout=10, decode='utf-8'):
    for tries in range(MAX_RETRY):
        try:
            with urllib.request.urlopen(html_url, timeout=timeout) as response:
                return response.read().decode(decode)
        except Exception as e:
            logging.warning(str(e) + ',html_url:{0}'.format(html_url))
            if tries < (MAX_RETRY - 1):
                continue
            else:
                print('Has tried {0} times to access url {1}, all failed!'.format(MAX_RETRY, html_url))
                return None

Answer 2

另一个答案不专门处理超时异常情况。要处理超时情况：

let video_div =  document.createElement('div');
video_div.id = 'video-div'
$(video_div).addClass('my-div')
$("body").append(($(video_div).draggable().resizable()))

let video_element;
video_element = document.createElement('video');
$(video_element).attr('id', 'my_video');
$(video_element).attr('class', 'video-js vjs-default-skin');
$(video_element).attr('width', '100%');
$(video_element).attr('height', '100%');
$(video_element).attr('controls', ' ');
$(video_element).attr('preload', 'auto');
$(video_element).attr('data-setup', '{}');

let source = document.createElement('source');
$(source).attr('type', "video/mp4");
$(source).attr('src', "http://grochtdreis.de/fuer-jsfiddle/video/sintel_trailer-480.mp4");

$(video_div).append(video_element)

$(video_element).append(source);

python - urllib.error.URLError：<urlopen error =“”timed =“”out =“”>

2 个答案: