如何在 XPath 1.0 中执行集合操作 [英] How to perform set operations in XPath 1.0

问题描述

我在 SO 和其他地方看到以下内容应该可以工作(这个例子直接取自 O'Reilly 的 XSLT Cookbook):

I've seen on SO and other places that the following is supposed to work (this example is lifted directly from O'Reilly's XSLT Cookbook):

(: intersection :)
$set1[count(. | $set2) = count($set2)]

(: difference :)
$set1[count(. | $set2) != count($set2)]

看起来应该没问题，但是当与实际路径而不是变量一起使用时，这似乎失败了.例如，给定以下文档

and it looks like it should be OK, however this seems to fail when used with actual paths rather than variables. For example, given the following document

<a>
  <new>
    <val>1</val>
    <val>2</val>
  </new>
  <old>
    <val>2</val>
    <val>3</val>
  </old>
</a>

和 XPath 表达式 /a/new/val[count(. |/a/old/val)=count(/a/old/val)]/text() 我期望获取节点集 { 2 } 而是获取 { 1 2 }.知道我做错了什么吗?

and the XPath expression /a/new/val[count(. | /a/old/val)=count(/a/old/val)]/text() I would expect to get the node-set { 2 } but instead am getting { 1 2 }. Any ideas what I'm doing wrong?

推荐答案

节点集交集的公式使用节点标识，而不是值标识.

The formulas for node-set intersection use node-identity, not value identity.

两个节点相同当且仅当 count($n1|$n2) =1

Two nodes are identical if and only if count($n1|$n2) =1

但是，您希望基于值标识相交.

However, you want to intersect based on value identity.

解决方案:

使用:

/a/new/val[. = /a/old/val]

这将选择任何 /a/new/val 至少存在一个 /a/old/val 元素，使得这两个元素的字符串值是一样.

this selects any /a/new/val for which there exists at least one /a/old/val element such that the string values of these two elements is the same.

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