How to get list of subelements that are only one level deep from Python3 xml.etree.Element?-CodePudding

If I have a Python3 xml.etree.Element (doc), is it possible to get a list (or iterable, or whatever) of child elements to that Element, but only one level deep? If so, how can I do this?

(Note: it appears the xml.etree.Element library is essentially the same for the version of Python I'm using (3.6.8) and the latest version (3.11).)

For example, say I have this xml:

<?xml version="1.0" encoding="UTF-8"?>
<Document>
  <name>myName</name>
  <SomeLevel1Element>
    <SomeLevel2Element>foo</SomeLevel2Element>
  </SomeLevel1Element>
  <SomeLevel1Element>
    <SomeLevel2Element>bar</SomeLevel2Element>
  </SomeLevel1Element>
  <AnotherLevel1Element>
  </AnotherLevel1Element>
</Document>

I want to do something like:

import xml.etree.ElementTree as ET
tree = ET.parse(PATH_TO_XML_FILE)
root = tree.getroot()
direct_children_of_root = <something>
for child in direct_children_of_root:
  print(child)

"""
<Element 'name' at 0x123abc>
<Element 'SomeLevel1Element' at 0x123bbc>
<Element 'SomeLevel1Element' at 0x123cbc>
<Element 'AnotherLevel1Element' at 0x123dbc>
"""

I cannot use findall(), find(), findtext(), since I may not know the amount or kinds of child elements of root.

I cannot use iter(), since that method appears to give all children of all children elements.

For example, using the same xml as above:

import xml.etree.ElementTree as ET
tree = ET.parse(PATH_TO_XML_FILE)
root = tree.getroot()
elems = root.iter()
for e in elems:
  print(e)

"""
<Element 'Document' at 0x123abc>
<Element 'name' at 0x123bbc>
<Element 'SomeLevel1Element' at 0x123cbc>
<Element 'SomeLevel2Element' at 0x123dbc>
<Element 'SomeLevel1Element' at 0x123ebc>
<Element 'SomeLevel2Element' at 0x123fbc>
<Element 'AnotherLevel1Element' at 0x124abc>
"""

CodePudding user response：

xml.etree.ElementTree has limited support for Xpath.
Using lxml instead

>>> from lxml import etree
>>> doc = etree.parse('tmp.xml')
>>> parentName = doc.getroot().tag
>>> level1 = doc.xpath(f'//*[parent::{parentName}]')
>>> for e in level1:
...     print(e.tag)
... 
name
SomeLevel1Element
SomeLevel1Element
AnotherLevel1Element

If xml.etree.ElementTree is required

>>> level1 = root.findall('.')[0].findall('./*')
>>> for e in level1:
...     print(e.tag)
... 
name
SomeLevel1Element
SomeLevel1Element
AnotherLevel1Element

CodePudding user response：

list(root) returns a list with the children of root.

direct_children_of_root = list(root)
for child in direct_children_of_root:
    print(child)

Even simpler is to just iterate over root directly.

for child in root:
    print(child)