Python:ElementTree，获取一个元素的命名空间字符串

Question

此 XML 文件名为 example.xml:

<?xml version="1.0"?>
<project xmlns="http://maven.apache.org/POM/4.0.0" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xsi:schemaLocation="http://maven.apache.org/POM/4.0.0 http://maven.apache.org/maven-v4_0_0.xsd">

  <modelVersion>14.0.0</modelVersion>
  <groupId>.com.foobar.flubber</groupId>
  <artifactId>uberportalconf</artifactId>
  <version>13-SNAPSHOT</version>
  <packaging>pom</packaging>
  <name>Environment for UberPortalConf</name>
  <description>This is the description</description>    
  <properties>
      <birduberportal.version>11</birduberportal.version>
      <promotiondevice.version>9</promotiondevice.version>
      <foobarportal.version>6</foobarportal.version>
      <eventuberdevice.version>2</eventuberdevice.version>
  </properties>
  <!-- A lot more here, but as it is irrelevant for the problem I have removed it -->
</project>

如果我加载 example.xml 并使用 ElementTree 解析它，我可以看到它的命名空间是 http://maven.apache.org/POM/4.0.0。

>>> from xml.etree import ElementTree
>>> tree = ElementTree.parse('example.xml')
>>> print tree.getroot()
<Element '{http://maven.apache.org/POM/4.0.0}project' at 0x26ee0f0>

我还没有找到一种方法来调用以仅从 Element 获取命名空间，而无需解析元素的 str(an_element)。看来一定有更好的办法。

Answer 1

对于 regular expression 来说，这是一项完美的任务。 .

import re

def namespace(element):
    m = re.match(r'\{.*\}', element.tag)
    return m.group(0) if m else ''