ElementTree XPath - Select Element based on attribute

Question

I am having trouble using the attribute XPath Selector in ElementTree, which I should be able to do according to the Documentation

Here's some sample code

XML

<root>
 <target name="1">
    <a></a>
    <b></b>
 </target>
 <target name="2">
    <a></a>
    <b></b>
 </target>
</root>

Python

def parse(document):
    root = et.parse(document)
    for target in root.findall("//target[@name='a']"):
        print target._children

I am receiving the following Exception:

expected path separator ([)

Ruphina

Using ElementTree 1.2.6, the attribute xpath features are only available in 1.3 and beyond. – Ruphina

Looks like findall only supports a subset XPath. See the mailing list discussion here. – Chopine

Why close this? It was useful for me... It is hardly off topic. – Stets

Answer 1

The syntax you're trying to use is new in ElementTree 1.3.

Such version is shipped with Python 2.7 or higher. If you have Python 2.6 or less you still have ElementTree 1.2.6 or less.

Coraleecoralie

Answer 2

There are several problems in this code.

1. Python's buildin ElementTree (ET for short) has no real XPATH support; only a limited subset By example, it doesn't support find-from-root expressions like //target.

   Notice: the documentation mentions "//", but only for children: So an expression as .//target is valid; //... is not!

   There is an alternative implementation: lxml which is more rich. It's seams that documentation is used, for the build-in code. That does not match/work.

2. The @name notation selects xml-attributes; the key=value expression within an xml-tag.

   So that name-value has to be 1 or 2 to select something in the given document. Or, one can search for targets with a child element 'a': target[a] (no @).

For the given document, parsed with the build-in ElementTree (v1.3) to root, the following code is correct and working:

• root.findall(".//target") Find both targets
• root.findall(".//target/a") Find two a-element
• root.findall(".//target[a]") This finds both target-element again, as both have an a-element
• root.findall(".//target[@name='1']") Find only the first target. Notice the quotes around 1 are needed; else a SyntaxError is raised
• root.findall(".//target[a][@name='1']") Also valid; to find that target
• root.findall(".//target[@name='1']/a") Finds only one a-element; ...

Huntingdon