How can we parse xml data that contains nodes with xml namespace tags in python?

Question

I am getting XML as a response so I want to parse it. I tried many python libraries but not get my desired results. So if you can help, it will be really appreciative.

The following code returns None:

xmlResponse = ET.fromstring(context.response_document)
a = xmlResponse.findall('.//Body')
print(a)

Sample XML Data:

<S:Envelope
       xmlns:S="http://www.w3.org/2003/05/soap-envelope">
       <S:Header>
           <wsa:Action s:mustUnderstand="1"
               xmlns:s="http://www.w3.org/2003/05/soap-envelope"
               xmlns:wsa="http://www.w3.org/2005/08/addressing">urn:ihe:iti:2007:RegistryStoredQueryResponse
           </wsa:Action>
       </S:Header>
       <S:Body>
           <query:AdhocQueryResponse status="urn:oasis:names:tc:ebxml-regrep:ResponseStatusType:Success"
               xmlns:query="urn:oasis:names:tc:ebxml-regrep:xsd:query:3.0">
               <rim:RegistryObjectList
                   xmlns:rim="u`enter code here`rn:oasis:names:tc:ebxml-regrep:xsd:rim:3.0"/>
               </query:AdhocQueryResponse>
           </S:Body>
       </S:Envelope>

I want to get status from it which is in Body. If you can suggest some changes of some library then please help me. Thanks

You can refer to: stackoverflow.com/questions/43589769/python3-parse-xml — Rithin Chalumuri
– Rithin Chalumuri, Commented Oct 31, 2019 at 20:10

Rithin Chalumuri · Accepted Answer · 2019-10-31 21:27:44Z

Given the following base code:

import xml.etree.ElementTree as ET

root = ET.fromstring(xml)

Let's build on top of it to get your desired output.

Your initial find for .//Body x-path returns NONE because it doesn't exist in your XML response.

Each tag in your XML has a namespace associated with it. More info on xml namespaces can be found here.

Consider the following line with xmlns value (xml-namespace):

<S:Envelope xmlns:S="http://www.w3.org/2003/05/soap-envelope">

The value of namespace S is set to be http://www.w3.org/2003/05/soap-envelope.

Replacing S in {S}Envelope with value set above will give you the resulting tag to find in your XML:

root.find('{http://www.w3.org/2003/05/soap-envelope}Envelope') #top most node

We would need to do the same for <S:Body>.

To get<S:Body> elements and it's child nodes you can do the following:

body_node = root.find('{http://www.w3.org/2003/05/soap-envelope}Body')

for response_child_node in list(body_node):
  print(response_child_node.tag) #tag of the child node
  print(response_child_node.get('status')) #the status you're looking for

Outputs:

{urn:oasis:names:tc:ebxml-regrep:xsd:query:3.0}AdhocQueryResponse
urn:oasis:names:tc:ebxml-regrep:ResponseStatusType:Success

Alternatively

You can also directly find all {query}AdhocQueryResponse in your XML using:

response_nodes = root.findall('.//{urn:oasis:names:tc:ebxml-regrep:xsd:query:3.0}AdhocQueryResponse')

for response in response_nodes:
  print(response.get('status'))

Outputs:

urn:oasis:names:tc:ebxml-regrep:ResponseStatusType:Success

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