在 XSLT 中按属性值对 XML 节点进行分组 [英] Grouping XML nodes by attribute value in XSLT

问题描述

我对 xslt 转换很陌生，我需要一种转换方面的帮助.我需要按其中一个属性对某种类型的所有节点进行分组，并列出每种属性的父节点.它是一种对文档中某些东西的用法的总结.我将展示简化示例.
输入:

Im quite new to xslt transforms and I need help with one kind of transformation. I need to group all nodes of certain type by one of it atributes and list parents of every kind of attribute. It is a kind of making summary of usage of certain things in the document. The I will present simplified example.
Input:

<root>
<node name="node1">
    <somechild child-id="1">
</node>
<node name="node2">
    <somechild child-id="2">
</node>
<node name="node3">
    <somechild child-id="1">
</node>
<node name="node4">
    <somechild child-id="2">
</node>
<node name="node5">
    <somechild child-id="3">
</node>
</root>

期望的输出:

<root>
<somechild child-id="1">
    <is-child-of>
        <node name="node1" />
        <node name="node3" />
    </is-child-of>
</somechild>
<somechild child-id="2">
    <is-child-of>
        <node name="node2" />
        <node name="node4" />
    </is-child-of>
</somechild>
<somechild child-id="3">
    <is-child-of>
        <node name="node5" />
    </is-child-of>
</somechild>
</root>

想法是，如果在许多节点中是相同的元素，则它们具有相同的 child-id.我需要找到每个 .我发现这个问题 XSLT 转换为 xml，按键分组有点相似，但开头有所有作者的声明，而我没有这样的声明，始终只是 .

Idea is that if is the same element in many nodes they have same child-id. I need to find all used by every . I found this question XSLT transformation to xml, grouping by key which is kind of similar but there is a declaration of all authors on the begining and I don't have such, is allways only a child of .

推荐答案

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

    <xsl:key name="k" match="somechild" use="@child-id"/>
    <xsl:key name="n" match="node" use="somechild/@child-id"/>

    <xsl:template match="root">
        <xsl:copy>
            <xsl:apply-templates 
                select="//somechild[generate-id(.) = generate-id(key('k', @child-id))]"/>
        </xsl:copy>
    </xsl:template>

    <xsl:template match="somechild">
        <xsl:copy>
            <xsl:apply-templates select="@*"/>

            <is-child-of>
                <xsl:apply-templates select="key('n', @child-id)"/>
            </is-child-of>
        </xsl:copy>

    </xsl:template>

    <xsl:template match="node">
        <xsl:copy>
            <xsl:apply-templates select="@*"/>
        </xsl:copy>
    </xsl:template>

    <xsl:template match="@*">
        <xsl:copy>
            <xsl:apply-templates select="@*"/>
        </xsl:copy>
    </xsl:template>

</xsl:stylesheet>

输出:

<root>
  <somechild child-id="1">
    <is-child-of>
      <node name="node1" />
      <node name="node3" />
    </is-child-of>
  </somechild>
  <somechild child-id="2">
    <is-child-of>
      <node name="node2" />
      <node name="node4" />
    </is-child-of>
  </somechild>
  <somechild child-id="3">
    <is-child-of>
      <node name="node5" />
    </is-child-of>
  </somechild>
</root>

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