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Modify xml file with extra namespace

I want to modify an existing xml file.
The layout of the existing file:
<?xml version="1.0" encoding="utf-8" standalone="yes"?>
<Document xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns="urn:iso:std:iso:20022:tech:xsd:pain.001.001.03">
After i modified a field in the xml i want to get the new xml file, but the modified file is different from the original.
<?xml version='1.0' encoding='UTF-8'?>
<Document xmlns="urn:iso:std:iso:20022:tech:xsd:pain.001.001.03">
Whats the difference between the files:
xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" //this is missing in the mutated file
So what i did:
ET.register_namespace('xsi', 'http://www.w3.org/2001/XMLSchema-instance')
ET.register_namespace('', "urn:iso:std:iso:20022:tech:xsd:pain.001.001.03")
#parse the data
tree = ET.parse(self.sepa_xml.path)
root = tree.getroot()
#add a subelement
body = ET.SubElement(root, "{http://www.w3.org/2001/XMLSchema-instance}")
The finale result:
<?xml version='1.0' encoding='UTF-8'?>
<Document xmlns="urn:iso:std:iso:20022:tech:xsd:pain.001.001.03" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance">
<CstmrCdtTrfInitn>
<PmtInf>
<PmtInfId>20220929085842-36645</PmtInfId>
<PmtMtd>TRF</PmtMtd>
<PmtTpInf>
<SvcLvl>
<Cd>SEPA</Cd>
</SvcLvl>
<CtgyPurp>
<Cd>SALA</Cd>
</CtgyPurp>
</PmtTpInf>
<ReqdExctnDt>2022-09-29</ReqdExctnDt>
<Dbtr>
<Nm>test name</Nm>
</Dbtr>
</DbtrAgt>
<ChrgBr>SLEV</ChrgBr>
<CdtTrfTxInf>
<PmtId>
<EndToEndId>20220929085842-36645/1</EndToEndId>
</PmtId>
</CdtTrfTxInf>
</PmtInf>
</CstmrCdtTrfInitn>
<xsi: /></Document>. // how can i delete this xsi tag ?
The problem is now that is get an extra tag at the end of the xml file:
<xsi: />. I assume this is because i added a subelement. How can ik delete this last tag ?

How to access the tag below another tag in xml using xml.dom.minidom in python?

I am using python 3.10.4 . I am new at parsing xml files.
like for eg, let the xml file be with the filename "test.xml":
<?xml version="1.0" encoding="UTF-8"?>
<tag1 name="1">
<tag2 name="a"></tag2>
</tag1>
<tag1 name = "2">
<tag2 name = "b"></tag2>
</tag1>
</xml>
python code
import xml.dom.minidom
file = xml.dom.minidom.parse('test.xml')
list = []
tags=file.getElementsByTagName("tag1")
for tag in tags:
if(tag.getAttribute("name")=="1"):
print(tag.getAttribute("tag2"))
So here I want to access the tag2 of tag1 with name="1". How can I do it?

Creating .docs file from .xml, .xslt and .docs template, using python

I am trying to create a .docx file, using the .docx template, .xml and .xslt file. I want to fill the placeholders in the .docx template file, with the data in the .xml file, and then generate a new word file, containing the data.
The template.docx file looks like this:
The data.xml file looks like this:
<root>
<person>
<Name>John</Name>
<profession>dentist</profession>
<city>Miami</city>
</person>
<person>
<Name>Mia</Name>
<profession>teacher</profession>
<city>London</city>
</person>
</root>
The parser.xslt file that I came up with looks like this:
<?xml version="1.0" encoding="UTF-8"?>
<xsl:stylesheet version="1.0" xmlns:xsl="">
<xsl:template match="/">
<xsl:for-each select="root/person">
<xsl:value-of select="Name"/>
<xsl:value-of select="profession"/>
<xsl:value-of select="city"/>
</xsl:for-each>
</xsl:template>
</xsl:stylesheet>
The output result.docs file should look like this:
My python code that I came up with looks like this:
import lxml.etree as ET
dom = ET.parse('data.xml')
xslt = ET.parse(r'parser.xslt')
transform = ET.XSLT(xslt)
newdom = transform(dom)
I don't know what the content of the .xslt file must be, in order to work, and how to create the result.docx
Any kind of help will be appreciable

Read a non formatted xml and export it again formatted? [duplicate]

Here is the code but the exported xml appears badly formatted.
import xml.etree.ElementTree as ET
import os
sampleXML = """<?xml version="1.0" encoding="ASCII"?>
<Metadata version="1.0">
<CODE_OK>510</CODE_OK>
<DeliveryDate>13/08/2018</DeliveryDate>
</Metadata>
"""
tree = ET.ElementTree(ET.fromstring(sampleXML))
for folder in os.listdir("YourPath"): #Iterate the dir
tree.find("CODE_OK").text = folder #Update dir name in XML
tree.write(open(os.path.join(r"Path", folder, "newxml.xml"), "wb")) #Write to XML
How to make the exported xml appear normally formatted?

I found in docs that xml module has an implementation of Document Object Model interface. I provide a simple example
from xml.dom.minidom import parseString
example = parseString(sampleXML) # your string
# write to file
with open('file.xml', 'w') as file:
example.writexml(file, indent='\n', addindent=' ')
Output:
<?xml version="1.0" ?>
<Metadata version="1.0">
<CODE_OK>510</CODE_OK>
<DeliveryDate>13/08/2018</DeliveryDate>
</Metadata>
Update
You can also write like this
example = parseString(sampleXML).toprettyxml()
with open('file.xml', 'w') as file:
file.write(example)
Output:
<?xml version="1.0" ?>
<Metadata version="1.0">
<CODE_OK>510</CODE_OK>
<DeliveryDate>13/08/2018</DeliveryDate>
</Metadata>
Update 2
I copy all your code and only add indent from this site. And for me is working correctly
import xml.etree.ElementTree as ET
import os
sampleXML = "your xml"
tree = ET.ElementTree(ET.fromstring(sampleXML))
indent(tree.getroot()) # this I add
for folder in os.listdir(path):
tree.find("CODE_OK").text = folder
tree.write(open(os.path.join(path, folder, "newxml.xml"), "wb"))

Reading an xml file for revision tag for a specified project

I have a sample.xml file as below..I need to scan for the "revision" tag for project "kernel/msm" and
print the word after "refs/heads"..i have a sample.xml file below and xml output?i can figure out the python part later,can any one provide inputs on how can this be done?
INPUT:-
Assume there is a variable project like below
project='kernel/msm'
sample.xml
<?xml version="1.0" encoding="utf-8"?>
<project name="platform/vendor/google/proprietary/code"
path="vendor/widevine"
revision="refs/heads/ab_mr"
x-grease-customer="none"
x-quic-dist="none"
x-ship="none" />
<!-- test Projects -->
<project name="kernel/msm"
path="kernel"
revision="refs/heads/msm-3.4"
x-grease-customer="none"
x-quic-dist="la"
x-ship="oss" />
......
EXPECTED OUTPUT:-
msm-3.4
Sample code:-
project='kernel/msm'
#open xml file
with open('./test.xml', 'r') as f:
#get the branch and project
for line in project :
if line in 'revision':
branch = line.split('/')[-1]
print branch
Thanks

import xml.etree.ElementTree as ET
import re
temp = 'refs/heads/'
name = 'kernel/msm'
pattern = re.compile('%s(.*)' % temp)
tree = ET.parse('sample.xml')
root = tree.getroot()
project = root.find("./project[#name='%s']" % name)
revision = project.get('revision')
res = pattern.match(revision)
print(res.group(1))
You have to wrap your xml data with a root node, for example <data> or it will raise a parse error.