Python 美丽的汤如何打印一个标签，而在它上面迭代

我的xml看起来像这样，我想得到位置

<?xml version="1.0" encoding="UTF-8"?>
<playlist version="1" xmlns="http://xspf.org/ns/0/">
 <trackList>
  <track>
   <location>file:///home/ashu/Music/Collections/randomPicks/ipod%20on%20sep%2009/Coldplay-Sparks.mp3</location>
   <title>Coldplay-Sparks</title>
  </track>
  <track>
   <location>file:///home/ashu/Music/Collections/randomPicks/gud%201s/Coldplay%20Warning%20sign.mp3</location>
   <title>Coldplay Warning sign</title>
  </track>....

但这不起作用，因为我得到：

AttributeError:'navigablesting'对象没有属性'location'

如何实现结果，提前感谢。

使用，速度更快，支持xpath:

>>> doc = lxml.etree.fromstring(yourxml)
>>> doc.xpath('//n:location/text()', namespaces={'n': 'http://xspf.org/ns/0/'})
['file:///home/ashu/Music/Collections/randomPicks/ipod%20on%20sep%2009/Coldplay-Sparks.mp3',
'file:///home/ashu/Music/Collections/randomPicks/gud%201s/Coldplay%20Warning%20sign.mp3']

---

您可以使用

findAll

：

>>> for track in soup.findAll('track'):
...     print track.title.string
...     print track.location.string
... 
Coldplay-Sparks
file:///home/ashu/Music/Collections/randomPicks/ipod%20on%20sep%2009/Coldplay-Sparks.mp3
Coldplay Warning sign
file:///home/ashu/Music/Collections/randomPicks/gud%201s/Coldplay%20Warning%20sign.mp3

---

是的，为了简洁起见我选了那个部分你想让我全部展示吗？是的。还要解释“不工作”是什么意思。findAll是否使用某种模式匹配？使用findAll似乎破坏了结构化数据库的整体意义document@Bunny：学习XPath足以解决80%的问题只需30分钟。nosklo，谢谢你的建议，我会防御性地学习，但可能在下个周末：P

>>> for track in soup.findAll('track'):
...     print track.title.string
...     print track.location.string
... 
Coldplay-Sparks
file:///home/ashu/Music/Collections/randomPicks/ipod%20on%20sep%2009/Coldplay-Sparks.mp3
Coldplay Warning sign
file:///home/ashu/Music/Collections/randomPicks/gud%201s/Coldplay%20Warning%20sign.mp3