I have this xml model.
link text
So I have to add some node (see the text commented) to this file.
How I can do it?
I have writed this partial code but it doesn't work:
xmldoc=minidom.parse(directory)
child = xmldoc.createElement("map")
for node in xmldoc.getElementsByTagName("Environment"):
node.appendChild(child)
Thanks in advance.
I downloaded your sample xml file and your code works fine. Your problem is most likely with the line: xmldoc=minidom.parse(directory), should this not be the path to the file you are trying to parse not to a directory? The parse() function parses an XML file it does not automatically parse all the XML files in a given directory.
If you change your code to something like below this should work fine:
xmldoc=minidom.parse("directory/model_template.xml")
child = xmldoc.createElement("map")
for node in xmldoc.getElementsByTagName("Environment"):
node.appendChild(child)
If you then execute the statement: print xmldoc.toxml() you will see that the map element has indeed been added to the Environment element: <Environment><map/></Environment>.
Related
I have an XML file that is when parsed reads a root different from the root I want it to read from.
the file starts with a tree that shows a summary of the data included in the file then another tree with the actual data in the subsequent section. However, when I parse it. it doesn't even read the tree that includes the data. It only reads the tree that includes the summary.
My question is how to modify my code to start reading the file from the tree that includes the data that I need
The XML file starts like that
I tried parsing it using xml.etree.ElementTree it reads the root "WorkflowSearch" what I want is for it to read the root at "SearchResults" so that I can read the fields into a pandas DataFrame
`tree = ET.parse('Workflows.xml')`
`root = tree.getroot()`
`root.tag, root.attrib`
I appreciate your help
I have tried to save the xml file in the following variable and later work on it as normal xml file. This is not working. How can I approach this situation. I need to edit the xml file without editing in the original file and without creating a new xml file. Is that possible?
comment_2 = open("cool.xml").read()
Thanks and Regards
You can use xml.etree.ElementTree to parse the XML file and then save it to a variable:
import xml.getElementTree as ET
tree = ET.parse('xml.etr.xml')
root = tree.getroot()
root.save(root)
root_variable = root_variable
Then you can save the xml file to an instance of ElementTree.
I am using xml.etree in python to generate an xml document, now while I am generating the xml document I want to get the xpath of a element that I have created, is there anyway of getting the absolute path while still creating the Element?
Below is an example of the generation process:
ParameterDeclaration = Element("ParameterDeclarations")
Parameter = Element("ParameterDeclaration")
Parameter.set("name", "Speed")
Parameter.set("value", action.speed)
Parameter.set("parameterType", "double")
ParameterDeclaration.append(Parameter)
Can I get the path of the Parameter element right after that last line?
Please note that this is only a small section of the xml file, the ParameterDeclaration is being appended to another element.
I am new to Python. I have installed signxml package and I am doing xml signature process.
Link to python package : https://pypi.org/project/signxml/
My xml file is getting generated. However XML signature code is little different. I was able to match most of the part but I donot have idea how to match following one.
Can any one please help me for that.
Different part is following tag
<Signature>
Above part should be like this one
<Signature xmlns="http://www.w3.org/2000/09/xmldsig#">
When i searched into signxml core file i found following note.
To specify the location of an enveloped signature within **data**, insert a
``<ds:Signature Id="placeholder"></ds:Signature>`` element in **data** (where
"ds" is the "http://www.w3.org/2000/09/xmldsig#" namespace). This element will
be replaced by the generated signature, and excised when generating the digest.
How to make change to get this changed.
Following is my python code
from lxml import etree
import xml.etree.ElementTree as ET
from signxml import XMLSigner, XMLVerifier
import signxml
el = ET.parse('example.xml')
root = el.getroot()
#cert = open("key/public.pem").read()
key = open("key/private.pem").read()
signed_root = XMLSigner(method=signxml.methods.enveloped,signature_algorithm='rsa-sha512',digest_algorithm="sha512").sign(root, key=key)
tree = ET.ElementTree(signed_root)
#dv = tree.findall(".//DigestValue");
#print(dv);
tree.write("new_signed_file.xml")
What above code doing is. It takes one xml file and do digital signature process and generates new file.
Can anyone please guide me where and what change should i do for this requirements ?
I am assuming you are using python signxml
Go to python setup and open this file Python\Lib\site-packages\signxml\ __init__.py
Open __init__.py file and do following changes.
Find following code
def _unpack(self, data, reference_uris):
sig_root = Element(ds_tag("Signature"), nsmap=self.namespaces)
Change with following code.
def _unpack(self, data, reference_uris):
#sig_root = Element(ds_tag("Signature"), nsmap=self.namespaces)
sig_root = Element(ds_tag("Signature"), xmlns="http://www.w3.org/2000/09/xmldsig#")
After doing this change re-compile your python signxml package.
Re-generate new xml signature file.
I am in a project in which I have the task to convert a text file to xml file.
But the restriction is, the file names should be same. I use element tree
module of python for this but while writing to the tree I use tree.write()
method in which I have to explicitly define / hardcode the xml filename
myself.
I want to know is there any procedure to automatically create the .xml file
with same name as the text file
the sh module provides great flexibility and may help you for what you want to do if I understand the question correctly. I have shown an example in the following:
import sh
name = "filename.xml"
new_name = name[-3] + "txt"
sh.touch(new_name)
from there you can open the created file and write directly to it.