Splitting XML file into multiple at given tags

Question

I want to split a XML file into multiple files. My workstation is very limited to Eclipse Mars with Xalan 2.7.1.

I can also use Python, but never used it before.

<?xml version="1.0" encoding="UTF-8"?>
<root>
    <row>
        <NAME>Doe</NAME>
        <FIRSTNAME>Jon</FIRSTNAME>
        <GENDER>M</GENDER>
    </row>
    <row>
        <NAME>Mustermann</NAME>
        <FIRSTNAME>Max</FIRSTNAME>
        <GENDER>M</GENDER>
    </row>
</root>

How can I transform them to look like this

<?xml version="1.0" encoding="UTF-8"?>
    <root>
        <row>
            <NAME>Doe</NAME>
            <FIRSTNAME>Jon</FIRSTNAME>
            <GENDER>M</GENDER>
        </row>
    </root>

I need every "row"-data in a single file with header. The data above is just an example. Most of the "row"-data has 16 attributes, but it varies from time to time.

Answer 1

Use Python ElementTree.

Create a file e.g. xmlsplitter.py. Add the code below (where file.xml is your xml file and assuming every row has a unique NAME element.).

import xml.etree.ElementTree as ET
context = ET.iterparse('file.xml', events=('end', ))
for event, elem in context:
    if elem.tag == 'row':
        title = elem.find('NAME').text
        filename = format(title + ".xml")
        with open(filename, 'wb') as f:
            f.write("<?xml version=\"1.0\" encoding=\"UTF-8\"?>\n")
            f.write(ET.tostring(elem))

Run it with

python xmlsplitter.py

Or if the names are not unique:

import xml.etree.ElementTree as ET
context = ET.iterparse('file.xml', events=('end', ))
index = 0
for event, elem in context:
    if elem.tag == 'row':
        index += 1
        filename = format(str(index) + ".xml")
        with open(filename, 'wb') as f:
            f.write("<?xml version=\"1.0\" encoding=\"UTF-8\"?>\n")
            f.write(ET.tostring(elem))