如何忽略 XPath 1.0 中的命名空间? [英] How to ignore namespaces in XPath 1.0?

问题描述

我在试图理解 XPath 的魔力时遇到了严重的问题.

I'm having serious issues trying to understand the magic that is XPath.

基本上，我有一些像这样的 XML:

Basically, I have some XML like so:

<a>
    <b>
        <c/>
    </b>
</a>

现在，我想计算我们有多少个 B，没有 C.这可以使用以下 XPath 轻松完成:

Now, I want to count how many B's we have, without C's. This can be done easily with the following XPath:

count(*/b[not(descendant::c)])

现在的问题很简单:我如何做同样的事情，同时忽略任何命名空间?

Now the question is this simple: How do I do the same thing, while ignoring any namespaces?

我会想象它是这样的吗?

I'd imagine it was something like this?

count(*/[local-name()='b']/[not(descendant::[local-name()='c'])])

但这并不正确.什么是我上面的等效 XPath 但忽略命名空间?

But this is not correct. What would be the equivalent XPath as I have above but that ignores namespaces?

推荐答案

给定的 XPath，

count(*/b[not(descendant::c)])

可以重写为忽略命名空间，如下所示:

can be re-written to ignore namespaces as follows:

count(*[local-name()='b' and not(descendant::*[local-name()='c'])])

请注意，通常最好不要破坏命名空间，而是通过分配命名空间前缀并在 XPath 表达式中使用它们来正确使用它们.

Note that it generally is better not to defeat namespaces but to work with them properly by assigning namespace prefixes and using them in your XPath expression.

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