Find path to the node using ElementTree - python-3.x

Question

Wih ElementTree, I can print every occurences of a specific tag (in my case ExpertSettingsSg
):
#!/usr/bin/env python3
import xml.etree.ElementTree as ET
root = ET.parse('mydoc.xml').getroot()
for children in root:
value=children.findall('.//ExpertSettingsSg')#tag I'm looking for
for settings in value:
if settings.text is not None:
print(settings.text)
But I didn't find a way to print the path of the occurence. Because my XML file has many levels and because ExpertSettingsSg can be almost at every level, I need to know where the ExpertSettingsSg come from. I'm looking for something like
Path to config xxxxxx = /root/xxx/aaaa/bbbb
If it's not possible with ElementTree, does any other library do the trick?
Thanks

Answer 1

If you already have the nodes, you can walk the tree and collect paths (borrowing the example from #valdi-bo):
from xml.etree import ElementTree as ET
txt ='''<main>
<x>
<a>
<ExpertSettingsSg id="1">x1</ExpertSettingsSg>
</a>
<b>
<dummy>xxxx</dummy>
</b>
</x>
<y>
<c>
<dummy>xxxx</dummy>
</c>
<d>
<ExpertSettingsSg id="2">x2</ExpertSettingsSg>
</d>
<e>
<ExpertSettingsSg id="3"/>
</e>
</y>
</main>'''
def node_walk(root: ET.Element):
path_to_node = []
node_stack = [root]
while node_stack:
node = node_stack[-1]
if path_to_node and node is path_to_node[-1]:
path_to_node.pop()
node_stack.pop()
yield (path_to_node, node)
else:
path_to_node.append(node)
for child in reversed(node):
node_stack.append(child)
root = ET.ElementTree(ET.fromstring(txt))
for node in root.findall('.//ExpertSettingsSg'):
for node_path, n in node_walk(root.getroot()):
if n is node:
xpath = "/".join(["."] + [n.tag for n in node_path[1:]] + [n.tag])
print(xpath, node)
# NOTE: Assert is to just show that the xpath is correct.
assert root.getroot().find(xpath) == node
You would get output like this:
./x/a/ExpertSettingsSg <Element 'ExpertSettingsSg' at 0x102cf5b80>
./y/d/ExpertSettingsSg <Element 'ExpertSettingsSg' at 0x102cf5db0>
./y/e/ExpertSettingsSg <Element 'ExpertSettingsSg' at 0x102cf5e50>
Instead of walking multiple times, we can walk once and collect all relevant nodes with path, like this:
xpaths = []
for node_path, n in node_walk(root.getroot()):
if n.tag == "ExpertSettingsSg":
xpath = "/".join(["."] + [n.tag for n in node_path[1:]] + [n.tag])
xpaths.append(xpath)
for xpath in xpaths:
node = root.getroot().find(xpath)
print(xpath, node)