Question

我在python中编写了以下代码，该代码转到数组中的url并找到有关该页面的特定信息 - 各种Web抓取工具。这个接受一个Reddit线程数组并输出每个线程的分数。该程序几乎从未完全执行。通常，在接收下面的错误消息之前，我将经历5次左右的迭代。有人可以帮我解决这个问题吗？

import urllib2
from bs4 import BeautifulSoup

urls = ['http://www.reddit.com/r/videos/comments/1i12o2/soap_precursor_to_a_lot_of_other_hilarious_shows/', 'http://www.reddit.com/r/videos/comments/1i12nx/kid_reporter_interviews_ryan_reynolds/', 'http://www.reddit.com/r/videos/comments/1i12ml/just_my_two_boys_going_full_derp_shocking_plot/']

for x in urls:
    f = urllib2.urlopen(x)
    data = f.read()
    soup = BeautifulSoup(data)
    span = soup.find('span', attrs={'class':'number'})
    print '{}:{}'.format(x, span.text)

我收到的错误消息是：

Traceback (most recent call last):
  File "C:/Users/jlazarus/Documents/YouTubeparse2.py", line 7, in <module>
    f = urllib2.urlopen(x)
  File "C:\Python27\lib\urllib2.py", line 127, in urlopen
    return _opener.open(url, data, timeout)
  File "C:\Python27\lib\urllib2.py", line 410, in open
    response = meth(req, response)
  File "C:\Python27\lib\urllib2.py", line 523, in http_response
    'http', request, response, code, msg, hdrs)
  File "C:\Python27\lib\urllib2.py", line 448, in error
    return self._call_chain(*args)
  File "C:\Python27\lib\urllib2.py", line 382, in _call_chain
    result = func(*args)
  File "C:\Python27\lib\urllib2.py", line 531, in http_error_default
    raise HTTPError(req.get_full_url(), code, msg, hdrs, fp)
HTTPError: HTTP Error 429: Unknown

Answer 1

使用try和except规则忽略以捕获错误，如果您只想跳过错误，这就是您想要的。

import urllib2
from bs4 import BeautifulSoup

urls = ['http://www.reddit.com/r/videos/comments/1i12o2/soap_precursor_to_a_lot_of_other_hilarious_shows/', 'http://www.reddit.com/r/videos/comments/1i12nx/kid_reporter_interviews_ryan_reynolds/', 'http://www.reddit.com/r/videos/comments/1i12ml/just_my_two_boys_going_full_derp_shocking_plot/']

for x in urls:
    try:
        f = urllib2.urlopen(x)
        data = f.read()
        soup = BeautifulSoup(data)
        span = soup.find('span', attrs={'class':'number'})
        print '{}:{}'.format(x, span.text)
    except HTTPError:
        print("HTTP Error, continuing")

请帮我解决这个错误消息？ - 蟒蛇

1 个答案: