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Table of Contents
Understanding XML Namespaces
Using ElementTree with Namespaces
Handling Default Namespaces
Alternative: Using lxml for Simpler Syntax
Home Backend Development XML/RSS Tutorial How to parse XML with namespaces in Python?

How to parse XML with namespaces in Python?

Oct 27, 2025 am 12:49 AM

Proper handling of namespace URIs is key to parsing namespaced XML. When using xml.etree.ElementTree, you need to define a namespace mapping, such as {'ns': '//m.sbmmt.com/global/link/99bbb17674e75c528140126a3deb730e'}, and pass it in through the namespaces parameter when searching for elements; you also need to specify a prefix for the default namespace; the lxml library provides a more concise syntax and supports XPath and curly braces to write URIs directly, improving flexibility.

How to parse XML with namespaces in Python?

When parsing XML with namespaces in Python, the key is to properly handle namespace prefixes and URIs. The xml.etree.ElementTree module (built-in) supports namespaces, but you need to include the namespace URI when searching for elements.

Understanding XML Namespaces

Namespaces prevent naming conflicts by qualifying element names with a URI. For example:


Value

Here, ns is a prefix bound to https://global.php.cn/link/99bbb17674e75c528140126a3deb730e . To access , you must use the full namespace URI.

Using ElementTree with Namespaces

Pass a namespace dictionary to find or findall methods:

  • Create a namespace map: {'ns': '//m.sbmmt.com/global/link/99bbb17674e75c528140126a3deb730e'}
  • Use it in queries: tree.find('ns:item', namespaces=ns_map)

Example:

import xml.etree.ElementTree as ET

xml_data = '''https://global.php.cn/link/99bbb17674e75c528140126a3deb730e"&gt ;
Apple
'''

root = ET.fromstring(xml_data)
ns = {'ns': ' https://global.php.cn/link/99bbb17674e75c528140126a3deb730e '}
item = root.find('ns:item', namespaces=ns)
print(item.text) #Output: Apple

Handling Default Namespaces

If the XML uses a default namespace (no prefix):


Banana

You still need to declare a prefix in your namespace map:

ns = {'df': 'http://example.com/default'}
item = root.find('df:item', namespaces=ns)

Alternative: Using lxml for Simpler Syntax

The lxml library offers more flexibility. Install with: pip install lxml

It supports XPath and cleaner namespace handling:

from lxml import etree

root = etree.fromstring(xml_data)
ns = {'ns': ' https://global.php.cn/link/99bbb17674e75c528140126a3deb730e '}
item = root.xpath('//ns:item', namespaces=ns)[0]
print(item.text)

lxml also allows using curly brace syntax: root.find('{https://global.php.cn/link/99bbb17674e75c528140126a3deb730e}item') .

Basically, always define the namespace URI when querying. Whether using ElementTree or lxml, accurate namespace mapping is essential for successful parsing.

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