Develop Reference

store content of a tag in a string with elementtree in python3

I'm using Python 3.7.2 and elementtree to copy the content of a tag in an XML file.
This is my XML file:
<?xml version="1.0" encoding="UTF-8"?>
<Document xmlns="urn:iso:std:iso:20022:tech:xsd:pain.001.003.03">
<CstmrCdtTrfInitn>
<GrpHdr>
<MsgId>nBblsUR-uH..6jmGgZNHLQAAAXgXN1Lu</MsgId>
<CreDtTm>2016-11-10T12:00:00.000+01:00</CreDtTm>
<NbOfTxs>1</NbOfTxs>
<CtrlSum>6</CtrlSum>
<InitgPty>
<Nm>TC 03000 Kunde 55 Protokollr ckf hrung</Nm>
</InitgPty>
</GrpHdr>
</CstmrCdtTrfInitn>
</Document>
I want to copy the content of the 'MsgId' tag and save it as a string.
I've manage to do this with minidom before, but due to new circumstances, I have to settle with elementtree for now.
This is that code with minidom:
dom = xml.dom.minidom.parse('H:\\app_python/in_spsh/{}'.format(filename_string))
message = dom.getElementsByTagName('MsgId')
for MsgId in message:
print(MsgId.firstChild.nodeValue)
Now I want to do the exact same thing with elementtree. How can I achieve this?

To get the text value of a single element, you can use the findtext() method. The namespace needs to be taken into account.
from xml.etree import ElementTree as ET
tree = ET.parse("test.xml") # Your XML document
msgid = tree.findtext('.//{urn:iso:std:iso:20022:tech:xsd:pain.001.003.03}MsgId')
With Python 3.8 and later, it is possible to use a wildcard for the namespace:
msgid = tree.findtext('.//{*}MsgId')

Change the key value in XML using Python

I need to modify the online_hostname key value in XML using python. I tried xml element tree but it does not work.
import xml.etree.ElementTree as ET
xml_tree = ET.parse('test.xml')
root = xml_tree.getroot()
root[0][0] = "requiredvalue"
test.xml file is as below:
<?xml version="1.0" encoding="UTF-8" ?>
<bzinfo>
<myidentity online_hostname="testdevice-air_2022_01_25"
bzlogin="me#abc.com" />
</bzinfo>
Error:
IndexError: child assignment index out of range

It is much better to explicitly search the required node (find will do in this case) instead of using indexes on the root node:
import xml.etree.ElementTree as ET
xml_tree = ET.parse('test.xml')
root = xml_tree.getroot()
myidentity_node = root.find('myidentity')
myidentity_node.attrib['online_hostname'] = 'required_value'
xml_tree.write('modified.xml')
modified.xml after running this code:
<bzinfo>
<myidentity bzlogin="me#abc.com" online_hostname="required_value" />
</bzinfo>

parsing XML file in python2.7

I know this is a very common question, but the kind of XML file and the kind of extraction of data i need is a little unique due to the nature of the xml file. So appreciate any help on the steps to extract the required data, with pyhton2.7
I have the below XML
<?xml version="1.0" encoding="UTF-8"?>
<Package xmlns="http://soap.sforce.com/2006/04/metadata">
<types>
<members>Mango.XYZ_DIG_Team_ABCDEF_Mango_Review</members>
<members>Mango.XYZ_DIG_Team_Reporting_Mango_Review</members>
<members>Opportunity.A_T_Occupier_City_Job_List</members>
<name>ListView</name>
</types>
<types>
<members>Modify_All_Data_Permission</members>
<members>Opportunity_Alerts_Implementation</members>
<members>Process_Builder_Permission</members>
<members>Regional_Business_Support</members>
<members>Reports_Dashboards_Data_Export_for_Super_Users</members>
<name>PermissionSet</name>
</types>
<types>
<members>SolutionManager</members>
<members>Standard</members>
<name>Profile</name>
</types>
<types>
<members>Mango.Set Verified Date and System Id</members>
<members>Mango.Update Mango Site With Billing Street%2C City%2C Country</members>
<members>Mango.Update Family Id on Mango when created</members>
<members>Opportunity.Set Opportunity Name</members>
<name>WorkflowRule</name>
</types>
<version>38.0</version>
</Package>
i am trying to extract only the members from the PermissionSet block. So that eventually i will have a file, that only have the entries like
Modify_All_Data_Permission
Opportunity_Alerts_Implementation
Process_Builder_Permission
Regional_Business_Support
Reports_Dashboards_Data_Export_for_Super_Users
I have been able to extract only the 'name' tag by
from xml.dom import minidom
doc = minidom.parse("path_to_xmlFile")
t = doc.getElementsByTagName("types")
for n in t:
name = n.getElementsByTagName("name")[0]
print name.firstChild.data
How can i extract the members and save that to a file?
Note: the number of 'members' are not fixed they varies.
I can also try with a different library, if it serves the purpose.

Probably easiest to use XPath
import xml.etree.ElementTree as ET
root = ET.parse('file.xml').getroot()
for member in root.findall(".//members/")
print(member.text)

This may help you!
import xml.etree.ElementTree as ET
tree = ET.parse('file.xml')
root = tree.getroot()
for data in root[1]:
print data.text

python ElementTree the text of element who has a child

When I try to read a text of a element who has a child, it gives None:
See the xml (say test.xml):
<?xml version="1.0"?>
<data>
<test><ref>MemoryRegion</ref> abcd</test>
</data>
and the python code that wants to read 'abcd':
import xml.etree.ElementTree as ET
tree = ET.parse('test.xml')
root = tree.getroot()
print root.find("test").text
When I run this python, it gives None, rather than abcd.
How can I read abcd under this condition?

Use Element.tail attribute:
>>> import xml.etree.ElementTree as ET
>>> tree = ET.parse('test.xml')
>>> root = tree.getroot()
>>> print root.find(".//ref").tail
abcd

ElementTree has a rather different view of XML that is more suited for nested data. .text is the data right after a start tag. .tail is the data right after an end tag. so you want:
print root.find('test/ref').tail

Accessing Elements with and without namespaces using lxml

Is there a way to search for the same element, at the same time, within a document that occur with and without namespaces using lxml? As an example, I would want to get all occurences of the element identifier irrespective of whether or not it is associated with a specific namespace. I am currently only able to access them separately as below.
Code:
from lxml import etree
xmlfile = etree.parse('xmlfile.xml')
root = xmlfile.getroot()
for l in root.iter('identifier'):
print l.text
for l in root.iter('{http://www.openarchives.org/OAI/2.0/provenance}identifier'):
print l.text
File: xmlfile.xml
<?xml version="1.0"?>
<record>
<header>
<identifier>identifier1</identifier>
<datestamp>datastamp1</datestamp>
<setSpec>setspec1</setSpec>
</header>
<metadata>
<oai_dc:dc xmlns:dc="http://purl.org/dc/elements/1.1/" xmlns:oai_dc="http://www.openarchives.org/OAI/2.0/oai_dc/" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xsi:schemaLocation="http://www.openarchives.org/OAI/2.0/oai_dc/ http://www.openarchives.org/OAI/2.0/oai_dc.xsd">
<dc:title>title1</dc:title>
<dc:title>title2</dc:title>
<dc:creator>creator1</dc:creator>
<dc:subject>subject1</dc:subject>
<dc:subject>subject2</dc:subject>
</oai_dc:dc>
</metadata>
<about>
<provenance xmlns="http://www.openarchives.org/OAI/2.0/provenance" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xsi:schemaLocation="http://www.openarchives.org/OAI/2.0/provenance http://www.openarchives.org/OAI/2.0/provenance.xsd">
<originDescription altered="false" harvestDate="2011-08-11T03:47:51Z">
<baseURL>baseURL1</baseURL>
<identifier>identifier3</identifier>
<datestamp>datestamp2</datestamp>
<metadataNamespace>xxxxx</metadataNamespace>
<originDescription altered="false" harvestDate="2010-10-10T06:15:53Z">
<baseURL>xxxxx</baseURL>
<identifier>identifier4</identifier>
<datestamp>2010-04-27T01:10:31Z</datestamp>
<metadataNamespace>xxxxx</metadataNamespace>
</originDescription>
</originDescription>
</provenance>
</about>
</record>

You could use XPath to solve that kind of issue:
from lxml import etree
xmlfile = etree.parse('xmlfile.xml')
identifier_nodes = xmlfile.xpath("//*[local-name() = 'identifier']")