没有名称空间的Python XpathEvaluator [英] Python XpathEvaluator without namespace

问题描述

我需要编写一个动态函数来查找ATOM xml文档的子树上的元素.

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

为此，我写了这样的话:

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)

上面的代码无法找到类别.

The code above fails to find category.

原因是getpath返回的XPath没有名称空间，而XPathEvaluator e()需要名称空间.

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

是否有一种方法可以使getpath返回路径中的名称空间，或者允许XPathEvaluator接受路径而不指定名称空间(或者以其他方式指定它)?

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)?

推荐答案

使用:

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

Or, if you want to be more precise:

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

或者简单地:

def:category

这篇关于没有名称空间的Python XpathEvaluator的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！