Python: ElementTree, get the namespace string of an Element

Question

This XML file is named 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> 

If I load example.xml and parse it with ElementTree I can see its namespace is 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> 

I have not found a method to call to get just the namespace from an Element without resorting to parsing the str(an_element) of an Element. It seems like there got to be a better way.

Answer 1

This is a perfect task for a regular expression.

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

Answer 2

The namespace should be in Element.tag right before the "actual" tag:

>>> root = tree.getroot() >>> root.tag '{http://maven.apache.org/POM/4.0.0}project' 

To know more about namespaces, take a look at ElementTree: Working with Namespaces and Qualified Names.