Python XML filtering using XPath [duplicate]

Question

I'm trying to get this thing working. I have a XML file and I need to filter the element 'title' using XPath. Afterwards I need to copy everything from under the C element to an external file, but that's not the point right now. I need to get this running using the xml.etree.cElementTree or xml.etree.ElementTree. I have already read a bunch of posts here on stackoverflow and also on other site's and got stuck. Soo.. First the XML structure:

<?xml version="1.0" encoding="UTF-8" standalone="yes"?>
<delivery xmlns="http://url" publicationdate="2013-08-28T09:10:32Z">
    <A>
        <B>
            <C>
                <Cid>XXXXXXXXX</Cid>
                <cref>111111-2222222</cref>
                <D>
                    <E/>
                    <F/>
                    <G/>
                    <H>
                        <Href>XXXXXXXXXXXX</Href>
                        <hcont name="XXXXXX" country="EN"/>
                    </H>
                    <I/>
                    <J/>
                    <K>XXXXXXXXX</K>
                    <oldK>XXXXXXX</oldK>
                    <title>
                        <content lang="en">TITLE</content>
                    </title>
                    <L>
                        <isL>false</isL>
                    </L>
                </D>
                <M>
                    <startTime>2013-08-28T03:00:00Z</startTime>
                    <endTime>2013-08-29T00:58:00Z</endTime>
                </M>
            </C>
        </B>
    </A>
</delivery>

I can't even get to find the Cid element by XPath. The script keeps returning 'None' or [] or just nothing.

import xml.etree.ElementTree as ET

doc = ET.ElementTree(file='short.xml') 
for x in doc.findall('./A/B/C'):
  print x.get('Cid').text

This one returns nothing. How to get this working? How to 'find' even the Cid element?

Community · Accepted Answer · 2017-05-23 12:11:07Z

3

You should pass namespaces argument to findall():

namespaces = {name_space_name_here: 'http://url'}
for x in doc.findall('./A/B/C', namespaces=namespaces):
    # do smth

Though, that won't work with a default namespace (just xmlns, as in your case).

In this case you can explicitly pass your namespace to the xpath:

for x in tree.findall('.//{%(uri)s}C' % {'uri': 'http://url'}):

Also see:

edited May 23, 2017 at 12:11

CommunityBot

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answered Sep 3, 2013 at 10:09

alecxe

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