Python XpathEvaluator without namespace

问题

I need to write a dynamic function that finds elements on a subtree of an ATOM xml document.

To do so, I've written something like this:

    tree = etree.parse(xmlFileUrl)
    e = etree.XPathEvaluator(tree, namespaces={'def':'http://www.w3.org/2005/Atom'})
    entries = e('//def:entry')
    for entry in entries:
        mypath = tree.getpath(entry) + "/category"
        category = e(mypath)

The code above fails to find category.

The reason is that getpath returns an XPath without namespaces, whereas the XPathEvaluator e() requires namespaces.

Is there a way to either make getpath return namespaces in the path, or allow XPathEvaluator to accept the path without specifying the namespace (or, rather, specifying it some other way)?

回答1:

Use:

*[local-name() = 'category']

Or, if you want to be more precise:

*[local-name() = 'category' and namespace-uri() = 'http://www.w3.org/2005/Atom']

Or simply:

def:category

来源：https://stackoverflow.com/questions/13091717/python-xpathevaluator-without-namespace