Parse XML File with nested Elements

Question

I'm trying export and xml file with nested elements that looks like this:

<?xml version="1.0" encoding="UTF-8"?>
<GetCategoriesResponse xmlns="urn:ebay:apis:eBLBaseComponents">
    <Timestamp>2015-11-16T03:51:35.809Z</Timestamp>
    <Ack>Success</Ack>
    <Version>927</Version>
    <Build>E927_CORE_API_17590338_R1</Build>
    <CategoryArray>
            <Category>
            <BestOfferEnabled>true</BestOfferEnabled>
            <AutoPayEnabled>true</AutoPayEnabled>
            <CategoryID>20081</CategoryID>
            <CategoryLevel>1</CategoryLevel>
            <CategoryName>Antiques</CategoryName>
            <CategoryParentID>20081</CategoryParentID>
            </Category>
    </CategoryArray>
    <CategoryCount>18282</CategoryCount>
    <UpdateTime>2015-09-01T22:57:09.000Z</UpdateTime>
    <CategoryVersion>113</CategoryVersion>
    <ReservePriceAllowed>true</ReservePriceAllowed>
    <MinimumReservePrice>0.0</MinimumReservePrice>
</GetCategoriesResponse>

I'm trying to acces the contents of the Tag with python element Tree I've tried different approaches such as:

import xml.etree.ElementTree as ET

xtree = ET.parse("categories.xml")
xroot = xtree.getroot()

for node in xroot.findall('CategoryArray'):
    for snode in node.findall('Category'):
            categoryName = snode.find("CategoryName").text
            categoryLevel = snode.find("CategoryLevel").text
            categoryParentID = snode.find('CategoryParentID').text
            
            print(categoryName, categoryLevel, categoryParentID)

but it does not output anything, what am I doing wrong?

simpleApp · Accepted Answer · 2022-01-10 04:31:48Z

3

Namespaces(ns) is key here, so you would need to factor in the ns. for more documentation pls refer.

import xml.etree.ElementTree as ET

xtree = ET.parse("categories.xml")
xroot = xtree.getroot()

ns = {'myns': 'urn:ebay:apis:eBLBaseComponents'} # define the namesapce

for node in xroot.findall('myns:CategoryArray',ns): # and then find them based on ns
    for snode in node.findall('myns:Category',ns):
            categoryName = snode.find("myns:CategoryName",ns).text
            categoryLevel = snode.find("myns:CategoryLevel",ns).text
            categoryParentID = snode.find('myns:CategoryParentID',ns).text
            
            print(categoryName, categoryLevel, categoryParentID)

answered Jan 10, 2022 at 4:31

simpleApp

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