How to edit xml config file in python 3? - python

I have a xml config file and needs to update particular attribute value.
<?xml version="1.0" encoding="utf-8"?>
<configuration>
<testCommnication>
<connection intervalInSeconds="50" versionUpdates="15"/>
</testCommnication>
</configuration>
I just need to update the "versionUpdates" value to "10".
How can i achieve this in python 3.
I have tried xml.etree and minidom and not able to achieve it.

Please use xml.etree.ElementTree to modify the xml:
Edit: If you want to retail the attribute order, use lxml instead. To install, use pip install lxml
# import xml.etree.ElementTree as ET
from lxml import etree as ET
tree = ET.parse('sample.xml')
root = tree.getroot()
# modifying an attribute
for elem in root.iter('connection'):
elem.set('versionUpdates', '10')
tree.write('modified.xml') # you can write 'sample.xml' as well
Content now in modified.xml:
<configuration>
<testCommnication>
<connection intervalInSeconds="50" versionUpdates="10" />
</testCommnication>
</configuration>

You can use xml.etree.ElementTree in Python 3 to handle XML :
import xml.etree.ElementTree
config_file = xml.etree.ElementTree.parse('your_file.xml')
config_file.findall(".//connection")[0].set('versionUpdates', 10))
config_file.write('your_new_file.xml')

Related

How can I retrieve specific information from a XML file using python?

I am working with Sentinel-2 Images, and I want to retrieve the Cloud_Coverage_Assessment from the XML file. I need to do this with Python.
Does anyone have any idea how to do this? I think I have to use the xml.etree.ElementTree but I'm not sure how?
The XML file:
<n1:Level-1C_User_Product xmlns:n1="https://psd-14.sentinel2.eo.esa.int/PSD/User_Product_Level-1C.xsd" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xsi:schemaLocation="https://psd-14.sentinel2.eo.esa.int/PSD/User_Product_Level-1C.xsd">
<n1:General_Info>
...
</n1:General_Info>
<n1:Geometric_Info>
...
</n1:Geometric_Info>
<n1:Auxiliary_Data_Info>
...
</n1:Auxiliary_Data_Info>
<n1:Quality_Indicators_Info>
<Cloud_Coverage_Assessment>90.7287</Cloud_Coverage_Assessment>
<Technical_Quality_Assessment>
...
</Technical_Quality_Assessment>
<Quality_Control_Checks>
...
</Quality_Control_Checks>
</n1:Quality_Indicators_Info>
</n1:Level-1C_User_Product>
read xml from file
import xml.etree.ElementTree as ET
tree = ET.parse('sentinel2.xml')
root = tree.getroot()
print(root.find('.//Cloud_Coverage_Assessment').text)
..and I want to retrieve the Cloud_Coverage_Assessment
Try the below (use xpath)
import xml.etree.ElementTree as ET
xml = '''<n1:Level-1C_User_Product xmlns:n1="https://psd-14.sentinel2.eo.esa.int/PSD/User_Product_Level-1C.xsd" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xsi:schemaLocation="https://psd-14.sentinel2.eo.esa.int/PSD/User_Product_Level-1C.xsd">
<n1:General_Info>
</n1:General_Info>
<n1:Geometric_Info>
</n1:Geometric_Info>
<n1:Auxiliary_Data_Info>
</n1:Auxiliary_Data_Info>
<n1:Quality_Indicators_Info>
<Cloud_Coverage_Assessment>90.7287</Cloud_Coverage_Assessment>
<Technical_Quality_Assessment>
</Technical_Quality_Assessment>
<Quality_Control_Checks>
</Quality_Control_Checks>
</n1:Quality_Indicators_Info>
</n1:Level-1C_User_Product>'''
root = ET.fromstring(xml)
print(root.find('.//Cloud_Coverage_Assessment').text)
output
90.7287

Script does not run correctly. No output

I'm writing a code that should read a XML file. But it doesn't run and I'm not able to find the problem. I think it has something to do with my laptop.
I've already tried to run it in the normal python3 shell and in there it works perfectly fine.
import os
import xml.etree.ElementTree as ET
tree = ET.parse('people.xml')
root = tree.getroot()
root[0].attrib
It should output this : {'name': 'Samy'} and it does in the python3 shell but it doesn't work in the script.
The XML file looks like this
<?xml version="1.0"?>
<PEOPLE>
<Person name="Samy">
<age>99</age>
<number>0176293747238</number>
</Person>
<Person name="Alkoholik">
<age>20</age>
<number>0176234923482</number>
</Person>
</PEOPLE>
adding the print() function gives you the wanted output.
import xml.etree.ElementTree as ET
XML = 'people.xml'
tree = ET.parse(XML)
root = tree.getroot()
print(root[0].attrib) #{'name': 'Samy'}
for i in root:
print(i.attrib) #{'name': 'Samy'}
#{'name': 'Alkoholik'}

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

How to get this value on xml file using python? (See the xml code)

This is my xml file. Settings.xml and This is what it inside have...
<settings>
<setting id="ipkey" value="1252473100" />
</settings>
How to get ipkey value using python? I think this understandable...
You can use the following code to help yourself get a particular attribute.
import xml.etree.ElementTree as ET
xml = ET.parse('yourfile.xml')
root = xml.getroot()
value = root.find('./setting').attrib['id']
For Example with lxml http://lxml.de/tutorial.html
>>> from lxml import etree
>>> tree = etree.XML("""<settings>
... <setting id="ipkey" value="1252473100" />
... </settings>""")
>>> tree.xpath("//setting[#id='ipkey']/#value")
['1252473100']

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']")

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