Modify a XML using ElementTree

Question

<grandParent>
    <parent>
       <child>Sam/Astronaut</child>
    </parent>
</grandParent>

I want to modify the above XML by adding another child tag inside parent tag. I'm doing something like this..

tree = ET.parse("test.xml")
a=ET.Element('parent')
b=ET.SubElement(a,"child")
b.text="Jay/Doctor"
tree.write("test.xml")

Is this the correct way of modifying the xml file? Any better way? or what else should I be taking care of in the above code?

Solution

Your code creates a whole new tree and adds Jay to it. You need to connect Jay to the existing tree, not to a new one.

Try this:

import xml.etree.ElementTree as ET

tree = ET.parse("test.xml")
a = tree.find('parent')          # Get parent node from EXISTING tree
b = ET.SubElement(a,"child")
b.text = "Jay/Doctor"
tree.write("test.xml")

If you want to search for a particular child, you could do this:

import xml.etree.ElementTree as ET
tree = ET.parse("test.xml")
a = tree.find('parent')
for b in a.findall('child'):
    if b.text.strip() == 'Jay/Doctor':
        break
else:
    ET.SubElement(a,"child").text="Jay/Doctor"
tree.write("test.xml")

Notice a.findall() (similar to a.find(), but returns all of the named elements). xml.etree has very limited search criteria. You might consider using lxml.etree and its .xpath() method.

Comments

[nick01]

thanks this works.I think I could use find() to verify no child similar to one I am about to enter exists. To avoid duplicates.

[nick01]

actually, how do I execute the find command inside the parent tag? Two parent tags could have same children but a single parent should not have duplicates..

[nick01]

that worked for me. This might be all I need from ElementTree. I will look into lxml.etree if I need more capability I guess. Thanks much!

[nick01]

Well I guess I spoke too soon - I am now having name spaces in the parent tag and my find all method now has a problem ... AttributeError: 'NoneType' object has no attribute 'findall' How do I handle this?

[nick01]

I am doing something like this. spaces={'xmlns':'maven.apache.org/POM/4.0.0', 'xsi':'w3.org/2001/XMLSchema-instance', 'schemaLocation':'maven.apache.org/xsd/maven-4.0.0.xsd'} using spaces in the find all method.. find all('module', namespaces=spaces)

[matth]

That sounds like a new question.

[nick01]

Here it is: stackoverflow.com/questions/25070180/…