I am trying to remove an element in an xml which contains a namespace.
Here is my code:
templateXml = """<?xml version="1.0" encoding="UTF-8"?>
<Metadata xmlns="http://www.amazon.com/UnboxMetadata/v1">
<Movie>
<CountryOfOrigin>US</CountryOfOrigin>
<TitleInfo>
<Title locale="en-GB">The Title</Title>
<Actor>
<ActorName locale="en-GB">XXX</ActorName>
<Character locale="en-GB">XXX</Character>
</Actor>
</TitleInfo>
</Movie>
</Metadata>"""
from lxml import etree
tree = etree.fromstring(templateXml)
namespaces = {'ns':'http://www.amazon.com/UnboxMetadata/v1'}
for checkActor in tree.xpath('//ns:Actor', namespaces=namespaces):
etree.strip_elements(tree, 'ns:Actor')
In my actual XML I have lots of tags, So I am trying to search for the Actor tags which contain XXX and completely remove that whole tag and its contents. But it's not working.
Use remove() method:
for checkActor in tree.xpath('//ns:Actor', namespaces=namespaces):
checkActor.getparent().remove(checkActor)
print etree.tostring(tree, pretty_print=True, xml_declaration=True)
prints:
<?xml version='1.0' encoding='ASCII'?>
<Metadata xmlns="http://www.amazon.com/UnboxMetadata/v1">
<Movie>
<CountryOfOrigin>US</CountryOfOrigin>
<TitleInfo>
<Title locale="en-GB">The Title</Title>
</TitleInfo>
</Movie>
</Metadata>
Related
I'd like to edit a KML file and remove all occurences of ExtendedData elements, wherever they are located in the file.
Here's the input XML file:
<?xml version="1.0" encoding="UTF-8"?>
<kml xmlns="http://earth.google.com/kml/2.2">
<Document>
<Style id="placemark-red">
<IconStyle>
<Icon>
<href>http://maps.me/placemarks/placemark-red.png</href>
</Icon>
</IconStyle>
</Style>
<name>My track</name>
<ExtendedData xmlns:mwm="https://maps.me">
<mwm:name>
<mwm:lang code="default">Blah</mwm:lang>
</mwm:name>
<mwm:lastModified>2020-04-05T14:17:18Z</mwm:lastModified>
</ExtendedData>
<Placemark>
<name></name>
…
<ExtendedData xmlns:mwm="https://maps.me">
<mwm:localId>0</mwm:localId>
<mwm:visibility>1</mwm:visibility>
</ExtendedData>
</Placemark>
</Document>
</kml>
And here's the code that 1) only removes the outermost occurence, and 2) requires adding the namespace to find it:
from lxml import etree
from pykml import parser
from pykml.factory import KML_ElementMaker as KML
with open("input.xml") as f:
doc = parser.parse(f)
root = doc.getroot()
ns = "{http://earth.google.com/kml/2.2}"
for pm in root.Document.getchildren():
#No way to get rid of namespace, for easier search?
if pm.tag==f"{ns}ExtendedData":
root.Document.remove(pm)
#How to remove innermost occurence of ExtendedData?
print(etree.tostring(doc, pretty_print=True))
Is there a way to remove all occurences in one go, or should I parse the whole tree?
Thank you.
Edit: The BeautifulSoup solution below requires adding an option "BeautifulSoup(my_xml,features="lxml")" to avoid the warning "No parser was explicitly specified".
Here's a solution using BeautifulSoup:
soup = BeautifulSoup(my_xml) # this is your xml
while True:
elem = soup.find("extendeddata")
if not elem:
break
elem.decompose()
Here's the output for your data:
<?xml version="1.0" encoding="UTF-8"?>
<html>
<body>
<kml xmlns="http://earth.google.com/kml/2.2">
<document>
<style id="placemark-red">
<IconStyle>
<Icon>
<href>http://maps.me/placemarks/placemark-red.png</href>
</Icon>
</IconStyle>
</style>
<name>
My track
</name>
<placemark>
<name>
</name>
</placemark>
</document>
</kml>
</body>
</html>
If you know the XML structure, try:
xml_root = ElementTree.parse(filename_path).getroot()
elem = xml_root.find('./ExtendedData')
xml_root.remove(elem)
or
xml_root = ElementTree.parse(filename_path).getroot()
p_elem = xml_root.find('/Placemark')
c_elem = xml_root.find('/Placemark/ExtendedData')
p_elem.remove(c_elem)
play with this ideas :)
if you don't know the xml structure, I think you need to parse the whole tree.
Simply run the empty template with Identity Transform using XSLT 1.0 which Python's lxml can run. No for/while loops or if logic needed. To handle the default namespace, define a prefix like doc:
XSLT (save a .xsl file, a special .xml file)
<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
xmlns:doc="http://earth.google.com/kml/2.2">
<xsl:output method="xml" omit-xml-declaration="yes" indent="yes"/>
<xsl:strip-space elements="*"/>
<!-- IDENTITY TRANSFORM -->
<xsl:template match="node()|#*">
<xsl:copy>
<xsl:apply-templates select="node()|#*"/>
</xsl:copy>
</xsl:template>
<!-- REMOVE ALL OCCURRENCES OF NODE -->
<xsl:template match="doc:ExtendedData"/>
</xsl:stylesheet>
Python
import lxml.etree as et
# LOAD XML AND XSL SOURCES
xml = et.parse('Input.xml')
xsl = et.parse('XSLT_Script.xsl')
# TRANSFORM INPUT
transform = et.XSLT(xsl)
result = transform(xml)
# PRINT TO SCREEN
print(result)
# SAVE TO FILE
with open('Output.kml', 'wb') as f:
f.write(result)
I working on a XML file that contains soap tags in it. I want to remove those soap tags as part of XML cleanup process.
How can I achieve it in either Python or Scala. Should not use shell script.
Sample Input :
<?xml version="1.0" encoding="UTF-8"?>
<soap:Envelope xmlns:soap="http://sample.com/">
<soap:Body>
<com:RESPONSE xmlns:com="http://sample.com/">
<Student>
<StudentID>100234</StudentID>
<Gender>Male</Gender>
<Surname>Robert</Surname>
<Firstname>Mathews</Firstname>
</Student>
</com:RESPONSE>
</soap:Body>
</soap:Envelope>
Expected Output :
<?xml version="1.0" encoding="UTF-8"?>
<com:RESPONSE xmlns:com="http://sample.com/">
<Student>
<StudentID>100234</StudentID>
<Gender>Male</Gender>
<Surname>Robert</Surname>
<Firstname>Mathews</Firstname>
</Student>
</com:RESPONSE>
This could help you!
from lxml import etree
doc = etree.parse('test.xml')
for ele in doc.xpath('//soap'):
parent = ele.getparent()
parent.remove(ele)
print(etree.tostring(doc))
I am getting this xml response, can anybody help me in getting the token from the xml tags?
<s:Envelope xmlns:s="http://schemas.xmlsoap.org/soap/envelope/"><s:Body><LoginResponse xmlns="http://videoos.net/2/XProtectCSServerCommand"><LoginResult xmlns:i="http://www.w3.org/2001/XMLSchema-instance"><RegistrationTime>2018-09-06T07:30:38.4571763Z</RegistrationTime><TimeToLive><MicroSeconds>3600000000</MicroSeconds></TimeToLive><TimeToLiveLimited>false</TimeToLiveLimited><Token>TOKEN#xxxxx#</Token></LoginResult></LoginResponse></s:Body></s:Envelope>
I have it as a string
Tried lxml and other libs too like ET but wasn't able to extract the token field. HELPPP
Update with a format xml to make you easy to read, FYI.
<?xml version="1.0" encoding="utf-8"?>
<s:Envelope xmlns:s="http://schemas.xmlsoap.org/soap/envelope/">
<s:Body>
<LoginResponse xmlns="http://videoos.net/2/XProtectCSServerCommand">
<LoginResult xmlns:i="http://www.w3.org/2001/XMLSchema-instance">
<RegistrationTime>2018-09-06T07:30:38.4571763Z</RegistrationTime>
<TimeToLive>
<MicroSeconds>3600000000</MicroSeconds>
</TimeToLive>
<TimeToLiveLimited>false</TimeToLiveLimited>
<Token>TOKEN#xxxxx#</Token>
</LoginResult>
</LoginResponse>
</s:Body>
</s:Envelope>
text = """
<?xml version="1.0" encoding="utf-8"?>
<s:Envelope xmlns:s="http://schemas.xmlsoap.org/soap/envelope/">
<s:Body>
<LoginResponse xmlns="http://videoos.net/2/XProtectCSServerCommand">
<LoginResult xmlns:i="http://www.w3.org/2001/XMLSchema-instance">
<RegistrationTime>2018-09-06T07:30:38.4571763Z</RegistrationTime>
<TimeToLive>
<MicroSeconds>3600000000</MicroSeconds>
</TimeToLive>
<TimeToLiveLimited>false</TimeToLiveLimited>
<Token>TOKEN#xxxxx#</Token>
</LoginResult>
</LoginResponse>
</s:Body>
</s:Envelope>
"""
from bs4 import BeautifulSoup
parser = BeautifulSoup(text,'xml')
for item in parser.find_all('Token'):
print(item.text)
Using lxml
Demo:
x = '''<?xml version="1.0" encoding="utf-8"?>
<s:Envelope xmlns:s="http://schemas.xmlsoap.org/soap/envelope/">
<s:Body>
<LoginResponse xmlns="http://videoos.net/2/XProtectCSServerCommand">
<LoginResult xmlns:i="http://www.w3.org/2001/XMLSchema-instance">
<RegistrationTime>2018-09-06T07:30:38.4571763Z</RegistrationTime>
<TimeToLive>
<MicroSeconds>3600000000</MicroSeconds>
</TimeToLive>
<TimeToLiveLimited>false</TimeToLiveLimited>
<Token>TOKEN#xxxxx#</Token>
</LoginResult>
</LoginResponse>
</s:Body>
</s:Envelope>'''
from lxml import etree
xmltree = etree.fromstring(x)
namespaces = {'content': "http://videoos.net/2/XProtectCSServerCommand"}
items = xmltree.xpath('//content:Token/text()', namespaces=namespaces)
print(items)
Output:
['TOKEN#xxxxx#']
What is the easiest way to navigate through XML with python?
<html>
<body>
<soapenv:envelope xmlns:soapenv="http://schemas.xmlsoap.org/soap/envelope/" xmlns:xsd="http://www.w3.org/2001/XMLSchema" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance">
<soapenv:body>
<getservicebyidresponse xmlns="http://www.something.com/soa/2.0/SRIManagement">
<code xmlns="">
0
</code>
<client xmlns="">
<action xsi:nil="true">
</action>
<actionmode xsi:nil="true">
</actionmode>
<clientid>
405965216
</clientid>
<firstname xsi:nil="true">
</firstname>
<id xsi:nil="true">
</id>
<lastname>
Last Name
</lastname>
<role xsi:nil="true">
</role>
<state xsi:nil="true">
</state>
</client>
</getservicebyidresponse>
</soapenv:body>
</soapenv:envelope>
</body>
</html>
I would go with regex and try to get the values of the lines I need but is there a pythonic way? something like xml[0][1] etc?
As #deceze already pointed out, you can use xml.etree.ElementTree here.
import xml.etree.ElementTree as ET
tree = ET.parse("path_to_xml_file")
root = tree.getroot()
You can iterate over all children nodes of root:
for child in root.iter():
if child.tag == 'clientid':
print(child.tag, child.text.strip())
Children are nested, and we can access specific child nodes by index, so root[0][1] should work (as long as the indices are correct).
I want parse this response from SOAP and extract text between <LoginResult> :
<?xml version="1.0" encoding="utf-8"?>
<soap:Envelope xmlns:soap="http://schemas.xmlsoap.org/soap/envelope/" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns:xsd="http://www.w3.org/2001/XMLSchema">
<soap:Body>
<LoginResponse xmlns="http://tempuri.org/wsSalesQuotation/Service1">
<LoginResult>45eeadF43423KKmP33</LoginResult>
</LoginResponse>
</soap:Body>
</soap:Envelope>
How I can do it using XML Python Libs?
import xml.etree.ElementTree as ET
tree = ET.parse('soap.xml')
print tree.find('.//{http://tempuri.org/wsSalesQuotation/Service1}LoginResult').text
>>45eeadF43423KKmP33
instead of print, do something useful to it.