python - 读取 XML 文件标签

Question

我想读取标签值，例如 <title> , <title_id>来自 xml 文件。 <title>的值读取成功。是否可以阅读 <title> , <title_id>具有相同的循环？
请帮助我，我是 XML 新手。

        <mediawiki xmlns="http://www.mediawiki.org/xml/export-0.5/" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xsi:schemaLocation="http://www.mediawiki.org/xml/export-0.5/ http://www.mediawiki.org/xml/export-0.5.xsd" version="0.5" xml:lang="en">
      <siteinfo>
        <sitename>Wiki</sitename>
        <case>first-letter</case>
        <namespaces>
          <namespace key="0" case="first-letter" />
        </namespaces>
      </siteinfo>
      <page>
        <title>Sex</title>
        <title_id>31239628</title_id>
        <revision>
          <id>437708703</id>
          <timestamp>2011-07-04T13:53:52Z</timestamp>
          <text xml:space="preserve" bytes="6830">{{ Hello}}

    </text>
        </revision>
      </page>
    </mediawiki>

我正在使用以下代码从文件中读取所有标题。而且工作正常。

import xml.etree.cElementTree as etree
tree = etree.parse('find_title.xml')
for value in tree.getiterator(tag='title'):
    print value.text

Answer 1

如果您要经常使用 XML，我建议您熟悉 XPATH .

这是使用我的首选 XML 库 lxml 的快速片段。

from lxml import etree

doc = etree.XML("""
<mediawiki xmlns="http://www.mediawiki.org/xml/export-0.5/" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xsi:schemaLocation="http://www.mediawiki.org/xml/export-0.5/ http://www.mediawiki.org/xml/export-0.5.xsd" version="0.5" xml:lang="en">
  <siteinfo>
    <sitename>Wiki</sitename>
    <case>first-letter</case>
    <namespaces>
      <namespace key="0" case="first-letter" />
    </namespaces>
  </siteinfo>
  <page>
    <title>Sex</title>
    <title_id>31239628</title_id>
    <revision>
      <id>437708703</id>
      <timestamp>2011-07-04T13:53:52Z</timestamp>
      <text xml:space="preserve" bytes="6830">{{ Hello}}
      </text>
    </revision>
  </page>
</mediawiki>
""")

def first(seq,default=None):
  for item in seq:
    return item
  return default

NSMAP=dict(mw="http://www.mediawiki.org/xml/export-0.5/")

print first(doc.xpath('/mw:mediawiki/mw:page/mw:title/text()',namespaces=NSMAP))
print first(doc.xpath('/mw:mediawiki/mw:page/mw:title_id/text()',namespaces=NSMAP))

产量:

Sex
31239628

---

Update - supposing multiple page elements

XPATH queries mostly return node sequences (hence the first function).

You could use a single query that returned the values of both tags for all of the pages. You would then have to group them together, if a subelement was missing from a page you'd be out of step. You could write the query to ensure the subelements existed, but you might want to know that there was a partial record, etc, etc.

So my first answer to this would be to loop through the pages like so:

for i,page in enumerate(doc.xpath('/mw:mediawiki/mw:page',namespaces=NSMAP)):
  title = first(page.xpath('./mw:title/text()',namespaces=NSMAP))
  title_id = first(page.xpath('./mw:title_id/text()',namespaces=NSMAP))
  print "Page %s: %s (%s)"  % (i,title,title_id)

产量:

Page 0: Sex (31239628)