从ElementTree findall返回的空列表

Question

我是xml解析和Python的新手，所以请耐心等待。我正在使用lxml来解析wiki转储，但我只想要每个页面，它的标题和文本。

现在我有了这个：

from xml.etree import ElementTree as etree

def parser(file_name):
    document = etree.parse(file_name)
    titles = document.findall('.//title')
    print titles

目前标题没有返回任何内容。我看过以前的答案：ElementTree findall() returning empty list和lxml文档，但大多数事情似乎都是针对解析HTML而定制的。

这是我的XML的一部分：

<mediawiki xmlns="http://www.mediawiki.org/xml/export-0.7/"     xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xsi:schemaLocation="http://www.mediawiki.org/xml/export-0.7/ http://www.mediawiki.org/xml/export-0.7.xsd" version="0.7" xml:lang="en">
  <siteinfo>
  <sitename>Wikipedia</sitename>
<base>http://en.wikipedia.org/wiki/Main_Page</base>
<generator>MediaWiki 1.20wmf9</generator>
<case>first-letter</case>
<namespaces>
  <namespace key="-2" case="first-letter">Media</namespace>
  <namespace key="-1" case="first-letter">Special</namespace>
  <namespace key="0" case="first-letter" />
  <namespace key="1" case="first-letter">Talk</namespace>
  <namespace key="2" case="first-letter">User</namespace>
  <namespace key="3" case="first-letter">User talk</namespace>
  <namespace key="4" case="first-letter">Wikipedia</namespace>
  <namespace key="5" case="first-letter">Wikipedia talk</namespace>
  <namespace key="6" case="first-letter">File</namespace>
  <namespace key="7" case="first-letter">File talk</namespace>
  <namespace key="8" case="first-letter">MediaWiki</namespace>
  <namespace key="9" case="first-letter">MediaWiki talk</namespace>
  <namespace key="10" case="first-letter">Template</namespace>
  <namespace key="11" case="first-letter">Template talk</namespace>
  <namespace key="12" case="first-letter">Help</namespace>
  <namespace key="13" case="first-letter">Help talk</namespace>
  <namespace key="14" case="first-letter">Category</namespace>
  <namespace key="15" case="first-letter">Category talk</namespace>
  <namespace key="100" case="first-letter">Portal</namespace>
  <namespace key="101" case="first-letter">Portal talk</namespace>
  <namespace key="108" case="first-letter">Book</namespace>
  <namespace key="109" case="first-letter">Book talk</namespace>
</namespaces>
  </siteinfo>
  <page>
    <title>Aratrum</title>
    <ns>0</ns>
    <id>65741</id>
    <revision>
  <id>349931990</id>
  <parentid>225434394</parentid>
  <timestamp>2010-03-15T02:55:02Z</timestamp>
  <contributor>
    <ip>143.105.193.119</ip>
  </contributor>
  <comment>/* Sources */</comment>
  <sha1>2zkdnl9nsd1fbopv0fpwu2j5gdf0haw</sha1>
  <text xml:space="preserve" bytes="1436">'''Aratrum''' is the Latin word for  [[plough]], and &quot;arotron&quot; (αροτρον) is the [[Greek language|Greek]] word. The   [[Ancient Greece|Greeks]] appear to have had diverse kinds of plough from the earliest  historical records. [[Hesiod]] advised the farmer to have always two ploughs, so that if  one broke the other might be ready for use. These ploughs should be of two kinds, the one  called &quot;autoguos&quot; (αυτογυος, &quot;self-limbed&quot;), in which the plough-tail  was of the same piece of timber as the share-beam and the pole; and the other called  &quot;pekton&quot; (πηκτον, &quot;fixed&quot;), because in it, three parts, which were of  three kinds of timber, were adjusted to one another, and fastened together by nails.

The ''autoguos'' plough was made from a [[sapling]] with two branches growing from its   trunk in opposite directions. In ploughing, the trunk served as the pole, one of the two     branches stood upwards and became the tail, and the other penetrated the ground and,    sometimes shod with bronze or iron, acted as the [[ploughshare]]. 

==Sources==
Based on an article from ''A Dictionary of Greek and Roman Antiquities,'' John Murray,     London, 1875.
ἄρατρον

==External links==
*[http://penelope.uchicago.edu/Thayer/E/Roman/Texts/secondary/SMIGRA*/Aratrum.html Smith's     Dictionary article], with diagrams, further details, sources.
[[Category:Agricultural machinery]]
[[Category:Ancient Greece]]
[[Category:Animal equipment]]</text>
</revision>
</page>

我也尝试过iterparse然后打印它找到的元素的标签：

for e in etree.iterparse(file_name):
    print e.tag

但它抱怨没有标签属性。

编辑：

Answer 1

问题是您没有考虑XML名称空间。 XML文档（及其中的所有元素）位于http://www.mediawiki.org/xml/export-0.7/命名空间中。要使其工作，您需要更改

titles = document.findall('.//title')

到

titles = document.findall('.//{http://www.mediawiki.org/xml/export-0.7/}title')

也可以通过namespaces参数提供命名空间：

NSMAP = {'mw':'http://www.mediawiki.org/xml/export-0.7/'}
titles = document.findall('.//mw:title', namespaces=NSMAP)

这适用于Python 2.7，但在Python 2.7 documentation中没有解释（Python 3.3 documentation更好）。

另请参阅http://effbot.org/zone/element-namespaces.htm和此SO问题，并附上答案：Parsing XML with namespace in Python via 'ElementTree'。

---

iterparse()的问题是由于此函数提供了(event, element)元组（而不仅仅是元素）。要获取标记名称，请更改

for e in etree.iterparse(file_name):
    print e.tag

到此：

for e in etree.iterparse(file_name):
    print e[1].tag

Answer 2

首先，您需要找到父元素page。我不知道这个嵌套有多少层，但一旦找到它，你就可以获得title标签：

>>> page_tag = ET.fromstring(xdata)
>>> title_tag = page_tag.find('title')
>>> title_tag.text
'Aratrum'

随着更多信息涌入，您可以这样做：

def parser(file_name):
    document = etree.parse(file_name)
    titles = []
    for page_tag in document.findall('page'):
        titles.append(page_tag.find('title').text)
    return titles

希望这有帮助！