Python ElementTree default namespace?

Question

Is there a way to define the default/unprefixed namespace in python ElementTree? This doesn't seem to work...

ns = {"":"http://maven.apache.org/POM/4.0.0"} pom = xml.etree.ElementTree.parse("pom.xml") print(pom.findall("version", ns)) 

Nor does this:

ns = {None:"http://maven.apache.org/POM/4.0.0"} pom = xml.etree.ElementTree.parse("pom.xml") print(pom.findall("version", ns)) 

This does, but then I have to prefix every element:

ns = {"mvn":"http://maven.apache.org/POM/4.0.0"} pom = xml.etree.ElementTree.parse("pom.xml") print(pom.findall("mvn:version", ns)) 

Using Python 3.5 on OSX.

EDIT: if the answer is "no", you can still get the bounty :-). I just want a definitive "no" from someone who's spent a lot of time using it.

Answer 1

NOTE: for Python 3.8+ please see this answer.

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There is no straight-forward way to handle the default namespaces transparently. Assigning the empty namespace a non-empty name is a common solution, as you've already mentioned:

ns = {"mvn":"http://maven.apache.org/POM/4.0.0"} pom = xml.etree.ElementTree.parse("pom.xml") print(pom.findall("mvn:version", ns)) 

Note that lxml.etree does not allow the use of empty namespaces explicitly. You would get:

ValueError: empty namespace prefix is not supported in ElementPath

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You can though, make things simpler, by removing the default namespace definition while loading the XML input data:

import xml.etree.ElementTree as ET import re   with open("pom.xml") as f:     xmlstring = f.read()   # Remove the default namespace definition (xmlns="http://some/namespace") xmlstring = re.sub(r'\sxmlns="[^"]+"', '', xmlstring, count=1)   pom = ET.fromstring(xmlstring)  print(pom.findall("version")) 

Answer 2

ElementTree in Python 3.8 allows empty string as a prefix, so you can declare:

ns = {'': 'http://maven.apache.org/POM/4.0.0'} 

and use that as the second arg in the find* methods.

Source: https://docs.python.org/3.8/library/xml.etree.elementtree.html?highlight=xml#xml.etree.ElementTree.Element.find