Find element by text with XPath in ElementTree

Question

Given an XML like the following:

<root>
    <element>A</element>
    <element>B</element>
</root>

How can I match the element with content A using ElementTree and its support for XPath? Thanks

Answer 1

AFAIK ElementTree does not support XPath. Has it changed?

Anyway, you can use lxml and the following XPath expression:

import lxml.etree
doc = lxml.etree.parse('t.xml')
print doc.xpath('//element[text()="A"]')[0].text
print doc.xpath('//element[text()="A"]')[0].tag

The result will be:

A
element

Answer 2

If you want to use the standard library ElementTree, rather than lxml, you can use iteration to find all sub elements with a particular text value. For example:

import sys
import xml.etree.ElementTree as etree

s = """<root>
    <element>A</element>
    <element>B</element>
</root>"""

e = etree.fromstring(s)

if sys.version_info < (2, 7):
    found = [element for element in e.getiterator() if element.text == 'A']
else:
    found = [element for element in e.iter() if element.text == 'A']

print found[0].text # This prints 'A', honestly!

Note: you may want to perform some stripping of the text value of your elements in the list comprehension.

Edit This will work to any depth in your XML tree. For example,

s = """<root>
    <element>A</element>
    <element><sub>A</sub></element>
</root>"""

found = [element for element in e.getiterator() if element.text == 'A']

for f in found:
    print f

will print

<Element element at 7f20a882e3f8>
<Element sub at 7f20a882e4d0>

Answer 3

You can use a subset of XPath in ElementTree. It isn't necessary to install any lib.

config.findall('.//*[element="A"]/element')

As the comment bellow from @Bionicegenius explains, the expression above just works if your element has no sibilings, but you get the idea.

It is possible to use XPath in ElementTree, and it is the simplest solution.