How to find a XML child element which has a default namespace in Python？

Question

My goal is to find the XML child element which has a default name.

XML:

<?xml version='1.0' encoding='UTF-8'?>
<all:config xmlns:all="urn:base:1.0">
    <interfaces xmlns="urn:ietf-interfaces">
        <interface>
          <name>eth0</name>
          <enabled>true</enabled>
          <ipv4 xmlns="urn:b-ip">
            <enabled>true</enabled>
          </ipv4>
          <tagging xmlns="urn:b:interfaces:1.0">true</tagging>
          <mac xmlns="urn:b:interfaces:1.0">00:00:10:00:00:11</mac>
        </interface>
  </interfaces>
</all:config>

I want to find the following element:

<mac xmlns="urn:b:interfaces:1.0">00:00:10:00:00:11</mac>

and change mac's text.

I have the following questions:

What is the xpath of mac?
How can I find "mac" using xpath since it has the default namespace?

My code does not work:

def set_element_value(file_name, element, new_value, order):
    filename = file_name
    tree = etree.parse(filename)
    root = tree.getroot()
    xml_string = etree.tostring(tree).decode('utf-8')
    my_own_namespace_mapping = {'prefix': 'urn:b:interfaces:1.0'}
    myele = root.xpath('.//prefix:mac', namespaces=my_own_namespace_mapping)
    myele[0].text = "aaa"
    for ele in root.xpath('.//prefix:mac', namespaces=my_own_namespace_mapping):
        if count_order == order:
            ele.text = str(new_value)
        count_order += 1
def main():
    filename ="./template/b.xml"
    element = ".//interfaces/interface/mac"
    new_value = "10"
    order = 0
    set_element_value(filename, element, new_value, order)
if __name__ == '__main__':
    main()

I tried to dig out in the stackoverflow, but no similar answer.

Could you please give me some tips? Thank you!

Jack · Accepted Answer · 2021-07-24 12:57:51Z

Thanks to Jack's methods, I fixed this issue:

The new code:

def set_element_value(file_name, element, new_value, order):
    filename = file_name
    tree = etree.parse(filename)
    tag_list = tree.xpath('.//*[local-name()="mac"]')
    print("tag:", tag_list, " and tag value:", tag_list[0].text)
    tag_list[0].text = "10"
    xml_string = etree.tostring(tree).decode('utf-8')
    print(xml_string)

def main():
    filename ="./template/b.xml"
    element = "mac"
    new_value = "10"
    order = 1
    set_element_value(filename, element, new_value, order)

if __name__ == '__main__':
    main()

output:

tag: [<Element {urn:ietf-interfaces}mac at 0x298fc4b69c0>]  and tag value: 10
<all:config xmlns:all="urn:base:1.0">
    <interfaces xmlns="urn:ietf-interfaces">
        <interface>
          <name>eth0</name>
          <enabled>true</enabled>
          <ipv4 xmlns="urn:b-ip">
            <enabled>true</enabled>
          </ipv4>
          <tagging xmlns="urn:b:interfaces:1.0">true</tagging>
          <mac xmln="urn:b:interfaces:1.0">10</mac>
        </interface>
  </interfaces>
</all:config>

Jack Fleeting · Accepted Answer · 2021-07-23 10:35:33Z

0

Your code seems to be a little too complicated than necessary. Try the following to get to the mac address:

ns = {"x":"urn:b:interfaces:1.0"}
root.xpath('//x:mac/text()',namespaces=ns)[0]

or if you don't want to deal with namespaces:

root.xpath('//*[local-name()="mac"]/text()')[0]

Output in either case is

00:00:10:00:00:11

answered Jul 23, 2021 at 10:35

Jack Fleeting

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1 Comment

Jack Over a year ago

Yes! Your methods work well! Really great thanks!

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