这简直要把我逼疯了,我已经为此挣扎了好几个小时了。任何帮助都将不胜感激。
我正在使用PyQuery 1.2.9 (它构建在lxml
之上)来抓取this URL。我只想获得.linkoutlist
部分中所有链接的列表。
这是我的完整请求:
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
但这会返回一个空数组。如果我使用下面的查询:
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&volume=19&issue=3&spage=125" ref="itool=Abstract&PrId=3159&uid=15107654&db=pubmed&log$=linkoutlink&nlmid=8609061" target="_blank">Lippincott Williams & Wilkins</a></li>
<li><a href="http://ovidsp.ovid.com/ovidweb.cgi?T=JS&PAGE=linkout&SEARCH=15107654.ui" ref="itool=Abstract&PrId=3682&uid=15107654&db=pubmed&log$=linkoutlink&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了带有大量<a>
标签的HTML.这似乎也是有效的HTML。
更多的实验表明,由于某些原因,lxml不喜欢打开div上的xmlns
属性。
我如何在lxml中忽略它,并像解析普通HTML一样解析它呢?
更新:正在尝试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)
发布于 2015-04-13 11:06:05
你需要处理命名空间,包括一个空的命名空间。
工作解决方案:
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,只需将parser
设置为"html"
并忽略名称空间:
links = pq('#maincontent .linkoutlist a', response.content, parser="html")
for link in links:
print link.attrib.get("title", "No title")
发布于 2015-04-13 08:21:55
祝你好运,让一个标准的XML/DOM解析器能够在大多数HTML上工作。最好的选择是使用BeautifulSoup (pip install beautifulsoup4
或easy_install beautifulsoup4
),它可以处理构建不正确的结构。也许就像这样呢?
import requests
from bs4 import BeautifulSoup
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')
bs = BeautifulSoup(response.content)
div = bs.find('div', class_='linkoutlist')
links = [ a['href'] for a in div.find_all('a') ]
>>> links
['http://meta.wkhealth.com/pt/pt-core/template-journal/lwwgateway/media/landingpage.htm?issn=0268-1315&volume=19&issue=3&spage=125', 'http://ovidsp.ovid.com/ovidweb.cgi?T=JS&PAGE=linkout&SEARCH=15107654.ui', 'https://www.researchgate.net/publication/e/pm/15107654?ln_t=p&ln_o=linkout', 'http://www.diseaseinfosearch.org/result/2199', 'http://www.nlm.nih.gov/medlineplus/antidepressants.html', 'http://toxnet.nlm.nih.gov/cgi-bin/sis/search/r?dbs+hsdb:@term+@rn+24219-97-4']
我知道这不是您想要使用的库,但在使用DOM时,我曾多次遇到麻烦。BeautifulSoup的创建者绕过了许多容易在野外发生的边缘情况。
发布于 2015-04-13 08:01:04
如果我没记错的话,不久前我自己也遇到过类似的问题。您可以通过将名称空间映射到None
来“忽略”它,如下所示:
sel = CSSSelector('#maincontent .rprt_all a', namespaces={None: "http://www.w3.org/1999/xhtml"})
https://stackoverflow.com/questions/29565335
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