I am trying to parse an xml and get the value of dir_path as below,however I dont seem to get the desired output,whats wrong here and how to fix it?
input.xml
<?xml version="1.0" ?>
<data>
<software>
<name>xyz</name>
<role>xyz</role>
<future>unknown</future>
</software>
<software>
<name>abc</name>
<role>abc</role>
<future>clear</future>
<dir_path cmm_root_path_var="COMP_softwareROOT">\\location\software\INR\</dir_path>
<loadit reduced="true">
<RW>yes</RW>
<readonly>R/</readonly>
</loadit>
<upload reduced="true">
</upload>
</software>
<software>
<name>def</name>
<role>def</role>
<future>clear</future>
<dir_path cmm_root_path_var="COMP2_softwareROOT">\\location1\software\INR\</dir_path>
<loadit reduced="true">
<RW>yes</RW>
<readonly>R/</readonly>
</loadit>
<upload reduced="true">
</upload>
</software>
</data>
CODE:-
tree = ET.parse(input.xml)
root = tree.getroot()
dir_path = root.find(".//dir_path")
print dir_path.text
OUTPUT:-
.\
EXPECTED OUTPUT:-
\\location\software\INR\
Try the following:
from xml.etree import ElementTree as ET
tree = ET.parse('filename.xml')
item = tree.find('software/[name="abc"]/dir_path')
print(item.text if item is not None else None)
Related
I'm trying to remove tags from an Xml.Alto file with remove.
My Alto file looks like this:
<alto xmlns="http://www.loc.gov/standards/alto/ns-v4#" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xsi:schemaLocation="http://www.loc.gov/standards/alto/ns-v4# http://www.loc.gov/standards/alto/v4/alto-4-2.xsd"> <Description>
<MeasurementUnit>pixel</MeasurementUnit>
<sourceImageInformation>
<fileName>filename</fileName>
</sourceImageInformation>
</Description>
<Layout>
<Page>
<PrintSpace>
<TextBlock>
<Shape><Polygon/></Shape>
<TextLine>
<Shape><Polygon/></Shape>
<String CONTENT="ABCDEF" HPOS="1234" VPOS="1234" WIDTH="1234" HEIGHT="1234" />
</TextLine>
</TextBlock>
</PrintSpace>
</Page>
</Layout>
</alto>
AND my code is :
import xml.etree.ElementTree as ET
tree = ET.parse("file.xml")
root = tree.getroot()
ns = {'alto': 'http://www.loc.gov/standards/alto/ns-v4#'}
ET.register_namespace("", "http://www.loc.gov/standards/alto/ns-v4#")
for Test in root.findall('.//alto:TextBlock', ns):
root.remove(Test)
tree.write('out.xml', encoding="UTF-8", xml_declaration=True)
Here is the error I get:
ValueError: list.remove(x): x not in list
Thanks a lot for your help 💐
ElementFather.remove(ElementChild) works only if the ElementChild is a sub-element of ElementFather. In your case, you have to call remove from PrintSpace.
import xml.etree.ElementTree as ET
tree = ET.parse("file.xml")
root = tree.getroot()
ns = {'alto': 'http://www.loc.gov/standards/alto/ns-v4#'}
ET.register_namespace("", "http://www.loc.gov/standards/alto/ns-v4#")
for Test in root.findall('.//alto:TextBlock', ns):
PrintSpace = root.find('.//alto:PrintSpace',ns)
PrintSpace.remove(Test)
tree.write('out.xml', encoding="UTF-8", xml_declaration=True)
Note: This code is only an example of a working solution, for sure you can improve it.
I have an xml file a shorter version is as follows:
<?xml version="1.0" encoding="UTF-8"?>
<DATA>
<_1>
<member_id>AFCE6DB97D4CD67D</member_id>
</_1>
<_2>
<member_id>AFCE6DB97D4CD67D</member_id>
</_2>
</DATA>
I am using the following code to parse
tree = ElementTree.parse(args['inputxml'])
root = tree.getroot()
for dat in root:
memberID = dat.find('member_id').text
I am able to parse the member id but not sure how to parse the serial number <_1>``<_2>etc. This number keeps extending with every new record in xml.
You could use xpath():
xml = """<?xml version="1.0" encoding="UTF-8"?>
<DATA>
<_1>
<member_id>AFCE6DB97D4CD67D</member_id>
</_1>
<_2>
<member_id>AFCE6DB97D4CD67D</member_id>
</_2>
</DATA>"""
root = etree.fromstring(xml)
members = root.xpath("//member_id")
for m in members:
print m.text, m.getparent().tag
This prints:
AFCE6DB97D4CD67D _1
AFCE6DB97D4CD67E _2
I have the following Input XML:
<?xml version="1.0" encoding="utf-8"?>
<Scenario xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xsi:noNamespaceSchemaLocation="Scenario.xsd">
<TestCase>test_startup_0029</TestCase>
<ShortDescription>Restart of the EVC with missing ODO5 board.</ShortDescription>
<Events>
<Event Num="1">Switch on the EVC</Event>
</Events>
<HW-configuration>
<ELBE5A>true</ELBE5A>
<ELBE5K>false</ELBE5K>
</HW-configuration>
<SystemFailure>true</SystemFailure>
</Scenario>
My Program does add three Tags to the XML but they are formatted false.
The Output XML looks like the following:
<Scenario xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xsi:noNamespaceSchemaLocation="Scenario.xsd">
<TestCase>test_startup_0029</TestCase>
<ShortDescription>Restart of the EVC with missing ODO5 board.</ShortDescription>
<Events>
<Event Num="1">Switch on the EVC</Event>
</Events>
<HW-configuration>
<ELBE5A>true</ELBE5A>
<ELBE5K>false</ELBE5K>
</HW-configuration>
<SystemFailure>true</SystemFailure>
<Duration>12</Duration><EVC-SW-Version>08.02.0001.0027</EVC-SW-Version><STAC-Release>08.02.0001.0027</STAC-Release></Scenario>
Thats my Source-Code:
class XmlManager:
#staticmethod
def write_xml(xml_path, duration, evc_sw_version):
xml_path = os.path.abspath(xml_path)
if os.path.isfile(xml_path) and xml_path.endswith(".xml"):
# parse XML into etree
root = etree.parse(xml_path).getroot()
# add tags
duration_tag = etree.SubElement(root, "Duration")
duration_tag.text = duration
sw_version_tag = etree.SubElement(root, "EVC-SW-Version")
sw_version_tag.text = evc_sw_version
stac_release = evc_sw_version
stac_release_tag = etree.SubElement(root, "STAC-Release")
stac_release_tag.text = stac_release
# write changes to the XML-file
tree = etree.ElementTree(root)
tree.write(xml_path, pretty_print=False)
else:
XmlManager.logger.log("Invalid path to XML-file")
def main():
xml = r".\Test_Input_Data_Base\blnmerf1_md1czjyc_REL_V_08.01.0001.000x\Test_startup_0029\Test_startup_0029.xml"
XmlManager.write_xml(xml, "12", "08.02.0001.0027")
My Question is how to add the new tags to the XML in the right format. I guess its working that way for parsing again the changed XML but its not nice formated. Any Ideas? Thanks in advance.
To ensure nice pretty-printed output, you need to do two things:
Parse the input file using an XMLParser object with remove_blank_text=True.
Write the output using pretty_print=True
Example:
from lxml import etree
parser = etree.XMLParser(remove_blank_text=True)
tree = etree.parse("Test_startup_0029.xml", parser)
root = tree.getroot()
duration_tag = etree.SubElement(root, "Duration")
duration_tag.text = "12"
sw_version_tag = etree.SubElement(root, "EVC-SW-Version")
sw_version_tag.text = "08.02.0001.0027"
stac_release_tag = etree.SubElement(root, "STAC-Release")
stac_release_tag.text = "08.02.0001.0027"
tree.write("output.xml", pretty_print=True)
Contents of output.xml:
<Scenario xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xsi:noNamespaceSchemaLocation="Scenario.xsd">
<TestCase>test_startup_0029</TestCase>
<ShortDescription>Restart of the EVC with missing ODO5 board.</ShortDescription>
<Events>
<Event Num="1">Switch on the EVC</Event>
</Events>
<HW-configuration>
<ELBE5A>true</ELBE5A>
<ELBE5K>false</ELBE5K>
</HW-configuration>
<SystemFailure>true</SystemFailure>
<Duration>12</Duration>
<EVC-SW-Version>08.02.0001.0027</EVC-SW-Version>
<STAC-Release>08.02.0001.0027</STAC-Release>
</Scenario>
See also http://lxml.de/FAQ.html#why-doesn-t-the-pretty-print-option-reformat-my-xml-output.
I'm a new to python so please bear with me as I try to explain what I am trying to do
here is my xml
<?xml version="1.0"?>
<playlist>
<list>
<txdate>2015-10-30</txdate>
<channel>cake</channel>
<name>Play List</name>
</list>
<eventlist>
<event type="MEDIA">
<title>title1</title>
<starttype>FIX</starttype>
<mediaid>a</mediaid>
<onairtime>2015-10-30T13:30:00:00</onairtime>
<som>00:00:40:03</som>
<duration>01:15:47:15</duration>
<reconcilekey>123</reconcilekey>
<category>PROGRAM</category>
<subtitles>
<cap>CLOSED</cap>
<file>a</file>
<lang>ENG</lang>
<lang>GER</lang>
</subtitles>
</event>
<event type="MEDIA">
<title>THREE DAYS AND A CHILD</title>
<mediaid>b</mediaid>
<onairtime>2015-10-30T14:45:47:15</onairtime>
<som>00:00:00:00</som>
<duration>01:19:41:07</duration>
<reconcilekey>321</reconcilekey>
<category>PROGRAM</category>
<subtitles>
<cap>CLOSED</cap>
<file>b</file>
<lang>ENG</lang>
<lang>GER</lang>
</subtitles>
</event>
</eventlist>
</playlist>
I would like to print all the mediaid values to a file
this is my code so far
import os
import xml.etree.ElementTree as ET
tree = ET.parse('data.xml')
root = tree.getroot()
wfile = 'new.csv'
for child in root:
child.find( "media type" )
for x in child.iter("mediaid"):
file = open(wfile, 'a')
file.write(str(x))
file.close
I tried this with a few other nonstandard libraries but I didn't have much success
For your requirement (as mentioned in the comments) -
just the mediaid from each <event type="MEDIA">
You should use findall() method of ElementTree to get all the event elements with type="MEDIA" , and then get the child mediaid element from it. Example -
import xml.etree.ElementTree as ET
tree = ET.parse('data.xml')
root = tree.getroot()
with open('new.csv','w') as outfile:
for elem in root.findall('.//event[#type="MEDIA"]'):
mediaidelem = elem.find('./mediaid')
if mediaidelem is not None:
outfile.write("{}\n".format(mediaidelem.text))
This is the sample data.
input.xml
<root>
<entry id="1">
<headword>go</headword>
<example>I <hw>go</hw> to school.</example>
</entry>
</root>
I'd like to put node and its descendants into . That is,
output.xml
<root>
<entry id="1">
<headword>go</headword>
<examplegrp>
<example>I <hw>go</hw> to school.</example>
</examplegrp>
</entry>
</root>
My poor and incomplete script is:
import codecs
import xml.etree.ElementTree as ET
fin = codecs.open(r'input.xml', 'rb', encoding='utf-8')
data = ET.parse(fin)
root = data.getroot()
example = root.find('.//example')
for elem in example.iter():
---and then I don't know what to do---
Here's an example of how it can be done:
text = """
<root>
<entry id="1">
<headword>go</headword>
<example>I <hw>go</hw> to school.</example>
</entry>
</root>
"""
import lxml.etree
import StringIO
data = lxml.etree.parse(StringIO.StringIO(text))
root = data.getroot()
for entry in root.xpath('//example/ancestor::entry[1]'):
examplegrp = lxml.etree.SubElement(entry,"examplegrp")
nodes = [node for node in entry.xpath('./example')]
for node in nodes:
entry.remove(node)
examplegrp.append(node)
print lxml.etree.tostring(root,pretty_print=True)
which will output:
<root>
<entry id="1">
<headword>go</headword>
<examplegrp><example>I <hw>go</hw> to school.</example>
</examplegrp></entry>
</root>
http://docs.python.org/3/library/xml.dom.html?highlight=xml#node-objects
http://docs.python.org/3/library/xml.dom.html?highlight=xml#document-objects
You probably want to follow some paradigm of creating a Document Element and appending reach result to it.
group = Document.createElement(tagName)
for found in founds:
group.appendNode(found)
Or something like this