title: How to parse next page by Beautiful Soup?
body: <p>I use code as below to parse page with next page:</p>
  
  <pre><code>def parseNextThemeUrl(url):
    ret = []
    ret1 = []
    html = urllib.request.urlopen(url)
    html = BeautifulSoup(html, PARSER)
    html = html.find('a', class_='pager_next')
    if html:
      html = urljoin(url, html.get('href'))
      ret1 = parseNextThemeUrl(html)
  
      for r in ret1:
          ret.append(r)
    else:
      ret.append(url)
    return ret
  </code></pre>
  
  <p>But I got error as below, how can I parse next link if there is a link.</p>
  
  <pre><code>Traceback (most recent call last):
  html = urllib.request.urlopen(url)
  File "/Library/Frameworks/Python.framework/Versions/3.5/lib/python3.5/urllib/request.py", line 162, in urlopen
  return opener.open(url, data, timeout)
  File "/Library/Frameworks/Python.framework/Versions/3.5/lib/python3.5/urllib/request.py", line 456, in open
  req.timeout = timeout
  AttributeError: 'list' object has no attribute 'timeout'
  </code></pre>