使用lxml来解析namepaced HTML？

Question

这让我完全疯了,我已经挣扎了好几个小时.任何帮助将非常感激.

我正在使用PyQuery 1.2.9(它构建在它之上lxml)来抓取这个URL.我只想获得该.linkoutlist部分中所有链接的列表.

这是我的全部要求:

response = requests.get('http://www.ncbi.nlm.nih.gov/pubmed/?term=The%20cost-effectiveness%20of%20mirtazapine%20versus%20paroxetine%20in%20treating%20people%20with%20depression%20in%20primary%20care')
doc = pq(response.content)
links = doc('#maincontent .linkoutlist a')
print links

但是返回一个空数组.如果我使用此查询:

links = doc('#maincontent .linkoutlist')

然后我得到这个HTML:

<div xmlns="http://www.w3.org/1999/xhtml" xmlns:xi="http://www.w3.org/2001/XInclude" class="linkoutlist">
   <h4>Full Text Sources</h4>
   <ul>
      <li><a title="Full text at publisher's site" href="http://meta.wkhealth.com/pt/pt-core/template-journal/lwwgateway/media/landingpage.htm?issn=0268-1315&amp;volume=19&amp;issue=3&amp;spage=125" ref="itool=Abstract&amp;PrId=3159&amp;uid=15107654&amp;db=pubmed&amp;log$=linkoutlink&amp;nlmid=8609061" target="_blank">Lippincott Williams &amp; Wilkins</a></li>
      <li><a href="http://ovidsp.ovid.com/ovidweb.cgi?T=JS&amp;PAGE=linkout&amp;SEARCH=15107654.ui" ref="itool=Abstract&amp;PrId=3682&amp;uid=15107654&amp;db=pubmed&amp;log$=linkoutlink&amp;nlmid=8609061" target="_blank">Ovid Technologies, Inc.</a></li>
   </ul>
   <h4>Other Literature Sources</h4>
   ...
</div>

所以父选择器确实返回带有大量<a>标签的HTML .这似乎也是有效的HTML.

更多的实验表明lxml xmlns由于某种原因不喜欢开放div上的属性.

我如何在lxml中忽略它,并像普通HTML一样解析它？

更新:尝试ns_clean,仍然失败:

    parser = etree.XMLParser(ns_clean=True)
    tree = etree.parse(StringIO(response.content), parser)
    sel = CSSSelector('#maincontent .rprt_all a')
    print sel(tree)

Answer 1

您需要处理名称空间,包括空名称空间.

工作方案:

from pyquery import PyQuery as pq
import requests


response = requests.get('http://www.ncbi.nlm.nih.gov/pubmed/?term=The%20cost-effectiveness%20of%20mirtazapine%20versus%20paroxetine%20in%20treating%20people%20with%20depression%20in%20primary%20care')

namespaces = {'xi': 'http://www.w3.org/2001/XInclude', 'test': 'http://www.w3.org/1999/xhtml'}
links = pq('#maincontent .linkoutlist test|a', response.content, namespaces=namespaces)
for link in links:
    print link.attrib.get("title", "No title")

打印与选择器匹配的所有链接的标题:

Full text at publisher's site
No title
Free resource
Free resource
Free resource
Free resource

或者,只设置parser到"html",而忘记了命名空间:

links = pq('#maincontent .linkoutlist a', response.content, parser="html")
for link in links:
    print link.attrib.get("title", "No title")