Retrieve attribute from XML node using namespace in Python

Question

I know that there are a few examples already of retrieving specific attributes from XML nodes, but I've not been successful in doing so with a namespace. I am able to retrieve an attribute without any namespaces, such as in this example.

Suppose I have the following "example.wsdl" file:

<?xml version="1.0" encoding="UTF-8"?><wsdl:definitions xmlns:wsdl="http://schemas.xmlsoap.org/wsdl/" name="Name of the file">
  <wsdl:documentation>Documentation of the file</wsdl:documentation>
</wsdl:definitions>

I'd like to retrieve the "name" attribute of the node "wsdl:defintions" I have tried doing the below:

from lxml import etree

tree = etree.parse("example.wsdl")
rootWSDL = tree.getroot()

print(tree.find('./wsdl:definitions', rootWSDL.nsmap).attrib['name'])

However, the above returns an empty list with the following message:

AttributeError: 'NoneType' object has no attribute 'attrib'

For what it's worth, the Python version I am using is 3.7.5

Rakesh · Accepted Answer · 2019-11-22 11:23:19Z

3

You can directly access it from the root.

Ex:

from lxml import etree

tree = etree.parse("example.wsdl")
rootWSDL = tree.getroot()
print(rootWSDL.attrib['name'])

#-->Name of the file

answered Nov 22, 2019 at 11:23

Rakesh

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3 Comments

Adam Over a year ago

Thank you. But suppose I have a use case where the node in question is not directly below the root? I'd like to find a way where I can explicitly specify which node I want to target rather than where it lies in the structure.

Rakesh Over a year ago

Use print(tree.find('wsdl:definitions', rootWSDL.nsmap).attrib['name']) without ./

mzjn Over a year ago

Adam: But suppose I have a use case where the node in question is not directly below the root? Why did you not ask about that use case in the first place then? @Rakesh did provide a reasonable answer. You have started a bounty, but the XML in the question has not been changed. What exactly are you asking?

Erik · Accepted Answer · 2019-12-02 02:14:51Z

2

+25

In your example you can't find the wsdl:definitions node with find because the rootWSDL is that node. If this is always the case just take rootWSDL.attrib['name'].

Your current xpath will only find elements that are direct children of rootWSDL. If your intent is to find some other namespaced node anywhere in the document simply make it a global xpath, i.e. replace:

'./wsdl:targetElementName'

with

'//wsdl:targetElementName'

answered Dec 2, 2019 at 2:14

Erik

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