Develop Reference

Python - replace root element of one xml file with another root element without its children

I have one xml file that looks like this, XML1:
<?xml version='1.0' encoding='utf-8'?>
<report>
</report>
And the other one that is like this,
XML2:
<?xml version='1.0' encoding='utf-8'?>
<report attrib1="blabla" attrib2="blabla" attrib3="blabla" attrib4="blabla" attrib5="blabla" >
<child1>
<child2>
....
</child2>
</child1>
</report>
I need to replace and put root element of XML2 without its children, so XML1 looks like this:
<?xml version='1.0' encoding='utf-8'?>
<report attrib1="blabla" attrib2="blabla" attrib3="blabla" attrib4="blabla" attrib5="blabla">
</report>
Currently my code looks like this but it won't remove children but put whole tree inside:
source_tree = ET.parse('XML2.xml')
source_root = source_tree.getroot()
report = source_root.findall('report')
for child in list(report):
report.remove(child)
source_tree.write('XML1.xml', encoding='utf-8', xml_declaration=True)
Anyone has ide how can I achieve this?
Thanks!

Try the below (just copy attrib)
import xml.etree.ElementTree as ET
xml1 = '''<?xml version='1.0' encoding='utf-8'?>
<report>
</report>'''
xml2 = '''<?xml version='1.0' encoding='utf-8'?>
<report attrib1="blabla" attrib2="blabla" attrib3="blabla" attrib4="blabla" attrib5="blabla" >
<child1>
<child2>
</child2>
</child1>
</report>'''
root1 = ET.fromstring(xml1)
root2 = ET.fromstring(xml2)
root1.attrib = root2.attrib
ET.dump(root1)
output
<report attrib1="blabla" attrib2="blabla" attrib3="blabla" attrib4="blabla" attrib5="blabla">
</report>

So here is working code:
source_tree = ET.parse('XML2.xml')
source_root = source_tree.getroot()
dest_tree = ET.parse('XML1.xml')
dest_root = dest_tree.getroot()
dest_root.attrib = source_root.attrib
dest_tree.write('XML1.xml', encoding='utf-8', xml_declaration=True)

Parsing XML in Python with ElementTree

I'm using the documentation here to try to get only the values (name,ip , netmask) for certain elements.
This is an example of the structure of my xml:
<?xml version="1.0" ?>
<rpc-reply xmlns="urn:ietf:params:xml:ns:netconf:base:1.0" xmlns:nc="urn:ietf:params:xml:ns:netconf:base:1.0" message-id="urn:uuid:5cf32451-91af-4f71-a0bd-ead244b81b1f">
<data>
<interfaces xmlns="urn:ietf:params:xml:ns:yang:ietf-interfaces">
<interface>
<name>GigabitEthernet1</name>
<type xmlns:ianaift="urn:ietf:params:xml:ns:yang:iana-if-type">ianaift:ethernetCsmacd</type>
<enabled>true</enabled>
<ipv4 xmlns="urn:ietf:params:xml:ns:yang:ietf-ip">
<address>
<ip>192.168.40.30</ip>
<netmask>255.255.255.0</netmask>
</address>
</ipv4>
<ipv6 xmlns="urn:ietf:params:xml:ns:yang:ietf-ip"/>
</interface>
<interface>
<name>GigabitEthernet2</name>
<type xmlns:ianaift="urn:ietf:params:xml:ns:yang:iana-if-type">ianaift:ethernetCsmacd</type>
<enabled>true</enabled>
<ipv4 xmlns="urn:ietf:params:xml:ns:yang:ietf-ip">
<address>
<ip>10.10.10.1</ip>
<netmask>255.255.255.0</netmask>
</address>
</ipv4>
<ipv6 xmlns="urn:ietf:params:xml:ns:yang:ietf-ip"/>
</interface>
</interfaces>
</data>
</rpc-reply>
Python code: This code returns nothing .
import xml.etree.ElementTree as ET
tree = ET.parse("C:\\Users\\Redha\\Documents\\test_network\\interface1234.xml")
root = tree.getroot()
namespaces = {'interfaces': 'urn:ietf:params:xml:ns:yang:ietf-interfaces' }
for elem in root.findall('.//interfaces:interfaces', namespaces):
s0 = elem.find('.//interfaces:name',namespaces)
name = s0.text
print(name)

interface = ET.parse('interface2.xml')
interface_root = interface.getroot()
for interface_attribute in interface_root[0][0]:
print(f"{interface_attribute[0].text}, {interface_attribute[3][0][0].text}, {interface_attribute[3][0][1].text}")

Fetch xml tag values recursively using ElementTree

I have an xmk of the type:
<SCHOOL>
<GROUP name="GetStudInfo">
<DATA>
<NAME type="char">Sahil Jha</NAME>
<STD>11th</STD>
</DATA>
<DATA>
<NAME type="char">Rashmi Kaur</NAME>
<STD>11th</STD>
</DATA>
<DATA>
<NAME type="char">Palak Bisht</NAME>
<STD>11th</STD>
</DATA>
</SCHOOL>
I need to fetch the values of NAME, STD.
I tried doing this:
e = ET.ElementTree(ET.fromstring(getunitinfo_str))
for elt in e.iter():
print("{} {}".format(elt.tag, elt.text))
But this was covering other values as well:
Output:
SCHOOL
GROUP
DATA
NAME Sahil Jha
STD 11th
DATA
NAME Rashmi Kaur
STD 11th
DATA
NAME Palak Bisht
STD 11th
{}
Expected O/p:
{'Sahil Jha':'11th', 'Rashmi Kaur'::'11th', 'Palak Bisht':'11th'}
But the formatting should be of the type NAME:STD. Where am I going wrong?

As mentionned by #furas you can use XPATH to find all DATA elements and then find
NAME and STD elements:
import xml.etree.ElementTree as ET
xml = '''<SCHOOL>
<GROUP name="GetStudInfo">
<DATA>
<NAME type="char">Sahil Jha</NAME>
<STD>11th</STD>
</DATA>
<DATA>
<NAME type="char">Rashmi Kaur</NAME>
<STD>11th</STD>
</DATA>
<DATA>
<NAME type="char">Palak Bisht</NAME>
<STD>11th</STD>
</DATA>
</GROUP>
</SCHOOL>'''
e = ET.fromstring(xml)
for data_tag in e.findall('DATA'):
name = data_tag.find('NAME')
std = data_tag.find('STD')
print("{} {}".format(name.text, std.text))
Or you can use a dict comprehension to get the dictionary you want:
my_dict = {
data_tag.find('NAME').text: data_tag.find('STD').text
for data_tag in e.findall('.//DATA')
}
print(my_dict)

You need something more then only print() - you need if/else to check elt.tag to get only NAME and `STD.
Because NAME and STD are different tags so you will have to remeber NAME in some variable to use it when you get STD
name = None # default value at start
for elt in e.iter():
if elt.tag == 'NAME':
name = elt # remember element
if elt.tag == 'STD':
print("{}:{}".format(name.text, elt.text))
Or you could use xpath like in #qouify answer.
Minimal working code
getunitinfo_str = '''
<SCHOOL>
<GROUP name="GetStudInfo">
<DATA>
<NAME type="char">Sahil Jha</NAME>
<STD>11th</STD>
</DATA>
<DATA>
<NAME type="char">Rashmi Kaur</NAME>
<STD>11th</STD>
</DATA>
<DATA>
<NAME type="char">Palak Bisht</NAME>
<STD>11th</STD>
</DATA>
</GROUP>
</SCHOOL>
'''
import xml.etree.ElementTree as ET
e = ET.ElementTree(ET.fromstring(getunitinfo_str))
name = None # to remeber element
for elt in e.iter():
if elt.tag == 'NAME':
name = elt
if elt.tag == 'STD':
print("{}:{}".format(name.text, elt.text))

One liner below
import xml.etree.ElementTree as ET
xml = '''<SCHOOL>
<GROUP name="GetStudInfo">
<DATA>
<NAME type="char">Sahil Jha</NAME>
<STD>11th</STD>
</DATA>
<DATA>
<NAME type="char">Rashmi Kaur</NAME>
<STD>116th</STD>
</DATA>
<DATA>
<NAME type="char">Palak Bisht</NAME>
<STD>17th</STD>
</DATA>
</GROUP>
</SCHOOL>'''
root = ET.fromstring(xml)
data = {x.find("NAME").text: x.find("STD").text for x in root.findall('.//DATA')}
print(data)
output
{'Sahil Jha': '11th', 'Rashmi Kaur': '116th', 'Palak Bisht': '17th'}

How to add space before and after CDATA in XML file

I want to create a function to modify XML content without changing the format. I managed to change the text but I can't do it without changing the format in XML.
So now, what I wanted to do is to add space before and after CDATA in a XML file.
Default XML file:
<?xml version="1.0" encoding="utf-8"?>
<Mapsxmlns="http://www.semi.org">
<Map>
<Device>
<ReferenceDevice/>
<Bin>
<Bin Bin="001"/>
</Bin>
<Data>
<Row> <![CDATA[001 001 001]]> </Row>
</Data>
</Device>
</Map>
</Maps>
And I am getting this result:
<?xml version="1.0" encoding="utf-8"?>
<Mapsxmlns="http://www.semi.org">
<Map>
<Device>
<ReferenceDevice/>
<Bin>
<Bin Bin="001"/>
</Bin>
<Data>
<Row><![CDATA[001 001 099]]></Row>
</Data>
</Device>
</Map>
</Maps>
However, I want the new xml to be like this:
<?xml version="1.0" encoding="utf-8"?>
<Mapsxmlns="http://www.semi.org">
<Map>
<Device>
<ReferenceDevice/>
<Bin>
<Bin Bin="001"/>
</Bin>
<Data>
<Row> <![CDATA[001 001 099]]> </Row>
</Data>
</Device>
</Map>
</Maps>
Here is my code:
from lxml import etree as ET
def xml_new(f,fpath,newtext,xmlrow):
xmlrow = 19
parser = ET.XMLParser(strip_cdata=False)
tree = ET.parse(f, parser)
root = tree.getroot()
for child in root:
value = child[0][2][xmlrow].text
text = ET.CDATA("001 001 099")
child[0][2][xmlrow] = ET.Element('Row')
child[0][2][xmlrow].text = text
child[0][2][xmlrow].tail = "\n"
ET.register_namespace('A', "http://www.semi.org")
tree.write(fpath,encoding='utf-8',xml_declaration=True)
return value
Anyone can help me on this? thanks in advance!

I don't quite understand what you want to do. Here's an example for you. I don't know if it can meet your needs.
from simplified_scrapy import SimplifiedDoc,req,utils
html ='''<?xml version="1.0" encoding="utf-8"?>
<Mapsxmlns="http://www.semi.org">
<Map>
<Device>
<ReferenceDevice/>
<Bin>
<Bin Bin="001"/>
</Bin>
<Data>
<Row> <![CDATA[001 001 001]]> </Row>
</Data>
</Device>
</Map>
</Maps>'''
doc = SimplifiedDoc(html)
row = doc.Data.Row # Get the node you want to modify.
row.setContent(" "+row.html+" ") # Modify the node content.
print (doc.html)
Result:
<?xml version="1.0" encoding="utf-8"?>
<Mapsxmlns="http://www.semi.org">
<Map>
<Device>
<ReferenceDevice />
<Bin>
<Bin Bin="001" />
</Bin>
<Data>
<Row> <![CDATA[001 001 001]]> </Row>
</Data>
</Device>
</Map>
</Maps>

thanks for all your help. I have found another way to achieve the result I want
This is the code:
# what you want to change
replaceby = '020]]> </Row>\n'
# row you want to change
row = 1
# col you want to change based on list
col = 3
file = open(file,'r')
line = file.readlines()
i = 0
editedXML=[]
for l in line:
if 'cdata' in l.lower():
i=i+1
if i == row:
oldVal = l.split(' ')
newVal = []
for index, old in enumerate(oldVal):
if index == col:
newVal.append(replaceby)
else:
newVal.append(old)
editedXML.append(' '.join(newVal))
else:
editedXML.append(l)
else:
editedXML.append(l)
file2 = open(newfile,'w')
file2.write(''.join(editedXML))
file2.close()

Sorting XML tags by child elements Python

I have a number of 'root' tags with children 'name'. I want to sort the 'root' blocks, ordered alphabetically by the 'name' element. Have tried lxml / etree / minidom but can't get it working...
I can't get it to parse the value inside the tags, and then sort the parent root tags.
<?xml version='1.0' encoding='UTF-8'?>
<roots>
<root>
<path>//1.1.1.100/Alex</path>
<name>Alex Space</name>
</root>
<root>
<path>//1.1.1.101/Steve</path>
<name>Steve Space</name>
</root>
<root>
<path>//1.1.1.150/Bethany</path>
<name>Bethanys</name>
</root>
</roots>
Here is what I have tried:
import xml.etree.ElementTree as ET
def sortchildrenby(parent, child):
parent[:] = sorted(parent, key=lambda child: child)
tree = ET.parse('data.xml')
root = tree.getroot()
sortchildrenby(root, 'name')
for child in root:
sortchildrenby(child, 'name')
tree.write('output.xml')

If you want to put the name nodes first:
x = """
<roots>
<root>
<path>//1.1.1.100/Alex</path>
<name>Alex Space</name>
</root>
<root>
<path>//1.1.1.101/Steve</path>
<name>Bethanys</name>
</root>
<root>
<path>//1.1.1.150/Bethany</path>
<name>Steve Space</name>
</root>
</roots>"""
import lxml.etree as et
tree = et.fromstring(x)
for r in tree.iter("root"):
r[:] = sorted(r, key=lambda ch: -(ch.tag == "name"))
print(et.tostring(tree).decode("utf-8"))
Which would give you:
<roots>
<root>
<name>Alex Space</name>
<path>//1.1.1.100/Alex</path>
</root>
<root>
<name>Bethanys</name>
<path>//1.1.1.101/Steve</path>
</root>
<root>
<name>Steve Space</name>
<path>//1.1.1.150/Bethany</path>
</root>
</roots>
But there is no need to sort if you just want to add them first, you can just remove and reinsert the name into index 0:
import lxml.etree as et
tree = et.fromstring(x)
for r in tree.iter("root"):
ch = r.find("name")
r.remove(ch)
r.insert(0, ch)
print(et.tostring(tree).decode("utf-8"))
If the nodes are actually not in sorted order and you want to rearrange the roots node alphabetically:
x = """
<roots>
<root>
<path>//1.1.1.100/Alex</path>
<name>Alex Space</name>
</root>
<root>
<path>//1.1.1.101/Steve</path>
<name>Steve Space</name>
</root>
<root>
<path>//1.1.1.150/Bethany</path>
<name>Bethanys</name>
</root>
</roots>"""
import lxml.etree as et
tree = et.fromstring(x)
tree[:] = sorted(tree, key=lambda ch: ch.xpath("name/text()"))
print(et.tostring(tree).decode("utf-8"))
Which would give you:
<roots>
<root>
<path>//1.1.1.100/Alex</path>
<name>Alex Space</name>
</root>
<root>
<path>//1.1.1.150/Bethany</path>
<name>Bethanys</name>
</root>
<root>
<path>//1.1.1.101/Steve</path>
<name>Steve Space</name>
</root>
</roots>
You can also combine with either of the first two approach two also rearrange the root nodes putting name first.

Try this:
import xml.etree.ElementTree as ET
xml="<?xml version='1.0' encoding='UTF-8'?><roots><root><path>//1.1.1.100/Alex</path><name>Alex Space</name></root><root><path>//1.1.1.101/Steve</path><name>Steve Space</name></root><root><path>//1.1.1.150/Bethany</path><name>Bethanys</name></root></roots>"
oldxml = ET.fromstring(xml)
names = []
for rootobj in oldxml.findall('root'):
names.append(rootobj.find('name').text)
newxml = ET.Element('roots')
for name in sorted(names):
for rootobj in oldxml.findall('root'):
if name == rootobj.find('name').text:
newxml.append(rootobj)
ET.dump(oldxml)
ET.dump(newxml)
I'm reading from a variable and dumpin it on screen.
You can change it read from file and dump it to a file like you need.