Question

我正在使用这个简单的代码

for l in bios:
    OpenThisLink = url + l
    response = urllib2.urlopen(OpenThisLink)

打开大约200个网址并用正则表达式（和BeautifulSoup）搜索它们，但是在十几个之后我得到这些错误并且IDLE退出。他们的意思是什么？我怎么处理它们？

谢谢。

Traceback (most recent call last):

  File "\PROJECTS\JD\jd10.py", line 15, in <module> response = urllib2.urlopen(OpenThisLink)

  File "C:\Python26\lib\urllib2.py", line 124, in urlopen return _opener.open(url, data, timeout)

  File "C:\Python26\lib\urllib2.py", line 389, in open response = meth(req, response)

  File "C:\Python26\lib\urllib2.py", line 502, in http_response 'http', request, response, code, msg, hdrs)

  File "C:\Python26\lib\urllib2.py", line 421, in error result = self._call_chain(*args)

  File "C:\Python26\lib\urllib2.py", line 361, in _call_chain result = func(*args)

  File "C:\Python26\lib\urllib2.py", line 597, in http_error_302 return self.parent.open(new)

  File "C:\Python26\lib\urllib2.py", line 389, in open response = meth(req, response)

  File "C:\Python26\lib\urllib2.py", line 502, in http_response 'http', request, response, code, msg, hdrs)

  File "C:\Python26\lib\urllib2.py", line 421, in error result = self._call_chain(*args)

  File "C:\Python26\lib\urllib2.py", line 361, in _call_chain result = func(*args)

  File "C:\Python26\lib\urllib2.py", line 597, in http_error_302 return self.parent.open(new)

  File "C:\Python26\lib\urllib2.py", line 389, in open response = meth(req, response)

  File "C:\Python26\lib\urllib2.py", line 502, in http_response 'http', request, response, code, msg, hdrs)

  File "C:\Python26\lib\urllib2.py", line 427, in error return self._call_chain(*args)

  File "C:\Python26\lib\urllib2.py", line 361, in _call_chain result = func(*args)

  File "C:\Python26\lib\urllib2.py", line 510, in http_error_default raise HTTPError(req.get_full_url(), code, msg, hdrs, fp) HTTPError: HTTP Error 404: Not Found

Answer 1

引发的错误是HTTPError - 具体而言，您的某个网址会引发404。你可以忽略它：

for l in bios:
    OpenThisLink = url + l
    try:
        response = urllib2.urlopen(OpenThisLink)
    except urllib2.HTTPError:
        pass

或者，您可以使用（稍微）更有意义的消息重新引发错误：

for l in bios:
    OpenThisLink = url + l
    try:
        response = urllib2.urlopen(OpenThisLink)
    except urllib2.HTTPError as e:
        raise Exception('Error opening %s: %s' % (e.geturl(), e))

Answer 2

我对你正在使用的特定库一无所知。但是，这对我来说就像一个大堆栈跟踪导致最终的原始错误：

HTTPError：HTTP错误404：未找到

我认为其中一个链接很糟糕，并触发了一个未被捕获的异常。

编辑：“坏”我的意思是服务器无法检索页面，因此出现404错误。

这些错误是什么，我该如何处理？

2 个答案: