Question

要转换的文档或多或少看起来像这样：

<?xml version="1.0" encoding="utf-8"?>
<root>
  <someCatalogProp>ąć</someCatalogProp>
  <meanProp>
           <node id="1">
    <someProperty>blabla1</someProperty>
    <children>
      <node idref="2"/>
    </children>
  </node>
  <node id="2">
    <someProperty>blabla2</someProperty>
    <children>
      <node idref="3"/>
    </children>
  </node>
  </meanProp>
  <node id="1">
    <someProperty>blabla1</someProperty>
    <children>
      <node idref="2"/>
    </children>
  </node>
  <node id="2">
    <someProperty>blabla2</someProperty>
    <children>
      <node idref="3"/>
    </children>
  </node>
  <node id="3">
    <someProperty>blabla3</someProperty>
    <children>
    </children>
  </node>
</root>

结果文档应如下所示：

<root>
<someCatalogProp>ąć</someCatalogProp>
    <node id = "1">
        <someProperty>blabla1</someProperty>
        <children>
             <node id = "2">
                 <someProperty>blabla2</someProperty>
                 <children>
                     <node id = "3">
                         <someProperty>blabla2</someProperty>
                         <children>
                         </children>
                     </node>
                </children>
            </node>
        </children>
    </node>
</root>

孩子的数量可以是多个。等级的深度不受限制。

转换xslt怎么样？提前谢谢。

Answer 1

使用 keys 来实现这一点非常简单。

如果您有格式正确的输入，例如：

<强> XML

<root>
    <node id="1">
        <someProperty>blabla1</someProperty>
        <children>
             <node idref="2"/>
        </children>
    </node>
    <node id="2">
        <someProperty>blabla2</someProperty>
        <children>
             <node idref="3"/>
        </children>
    </node>
    <node id="3">
        <someProperty>blabla2</someProperty>
        <children>
        </children>
    </node>
</root>

应用以下样式表：

XSLT 1.0

<xsl:stylesheet version="1.0" 
xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
<xsl:output method="xml" version="1.0" encoding="UTF-8" indent="yes"/>
<xsl:strip-space elements="*"/>

<xsl:key name="child" match="node" use="@id" />
<xsl:key name="parent" match="node" use="@idref" />

<!-- identity transform -->
<xsl:template match="@*|node()">
    <xsl:copy>
        <xsl:apply-templates select="@*|node()"/>
    </xsl:copy>
</xsl:template>

<xsl:template match="/root">
    <xsl:copy>
        <xsl:apply-templates select="node[not(key('parent', @id))]"/>
    </xsl:copy>
</xsl:template>

<xsl:template match="node[@idref]">
    <xsl:apply-templates select="key('child', @idref)"/>
</xsl:template>

</xsl:stylesheet>

将产生：

<强>结果

<?xml version="1.0" encoding="UTF-8"?>
<root>
   <node id="1">
      <someProperty>blabla1</someProperty>
      <children>
         <node id="2">
            <someProperty>blabla2</someProperty>
            <children>
               <node id="3">
                  <someProperty>blabla2</someProperty>
                  <children/>
               </node>
            </children>
         </node>
      </children>
   </node>
</root>

转换将具有相同名称的节点移动到其父节点中

1 个答案: