XPath 1.0 在 XML 树中具有属性的最接近的前一个和/或祖先节点 [英] XPath 1.0 closest preceding and/or ancestor node with an attribute in a XML Tree

问题描述

这里是三个 XML 树

Here are three XML-trees

(1)

<?xml version="1.0" encoding="UTF-8"?>
<content>
   <section id="1">
        <section id="2"/>
        <section id="3"/>
        <section id="9"/>
    </section>
    <section id="4">
        <section id="5">
            <section>
                <bookmark/> <!-- here's the bookmark-->
                <section id="6">
                    <section id="7">
                        <section id="8"/>
                    </section>
               </section>
            </section>
        </section>
    </section>
</content>

(2)

<?xml version="1.0" encoding="UTF-8"?>
<content>
    <section id="1"/>
    <section id="2">
        <section id="9">
            <section id="10">
                <section id="11"/>
            </section>            
        </section>
        <section>
            <section id="4">
                <section id="5"/>
            </section>            
        </section>
        <section/>
        <bookmark/> <!-- here's the bookmark-->
        <section id="6">
            <section id="7">
                <section id="8"/>
            </section>
        </section>
    </section>
</content>

在这两种情况下，所需的结果都是 id 5.

The desired result is in both cases the id 5.

使用 XSLT 1.0 和 XPath 1.0 我可以从 (1) 获取祖先

With XSLT 1.0 and XPath 1.0 I can either get the ancestor from (1)

<xsl:value-of select="//bookmark/ancestor::*[@id][1]/@id"/>

或 (2) 中的前一个节点

or the preceding node from (2)

<xsl:value-of select="//bookmark/preceding::*[@id][1]/@id"/>

如何使用书签中的 ID 获取最近祖先或前一个节点?
我需要一个 xsl:value-of 匹配这两种情况.谢谢.

How do I get the nearest ancestor or preceding node with an id from my bookmark?
I need a single xsl:value-of which matches both cases. Thanks.

解决方案还应涵盖此结构.所需的 ID 仍然是 5.

The solution should also cover this structure. Desired id is still 5.

(3)

<?xml version="1.0" encoding="UTF-8"?>
<content>
   <section id="1">
        <section id="2"/>
        <section id="3"/>
        <section id="9"/>
    </section>
    <section id="4">
        <section>
            <section id="10"/>
            <section id="5"/>
            <section>
                <bookmark/> <!-- here's the bookmark-->
                <section id="6">
                    <section id="7">
                        <section id="8"/>
                    </section>
                </section>
            </section>
        </section>
    </section>
</content>

推荐答案

使用:

    (//bookmark/ancestor::*[@id][1]/@id 
| 
    //bookmark/preceding::*[@id][1]/@id
     )
     [last()]

验证:使用 XSLT 作为 XPath 的宿主，进行以下转换:

Verification: Using XSLT as host of XPath, the following transformation:

<xsl:stylesheet version="1.0"
 xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
 <xsl:output omit-xml-declaration="yes" indent="yes"/>

 <xsl:template match="/">
  <xsl:value-of select=
  "(//bookmark/ancestor::*[@id][1]/@id
  |
   //bookmark/preceding::*[@id][1]/@id
   )
    [last()]"/>
 </xsl:template>
</xsl:stylesheet>

应用于所提供的三个 XML 文档中的任何一个时，都会产生所需的正确结果:

5

我强烈建议使用 XPath Visualizer 来玩/学习 XPath.

I strongly recomment using the XPath Visualizer for playing with / learning XPath.

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