需求：使用高级XSLT实现XML到XML的转换（附示例XML）

Got it, let's work through this. Your input XML has a <root> containing multiple <name> nodes, each with <elements> tagged with a type (T attribute) and sequence number (S attribute). Since you didn't specify the exact target XML structure you need, I'll walk through a common, flexible transformation that groups elements by their T attribute within each <name>—this is easy to adapt to your specific output requirements.

Input XML #

<root>
  <name>
    <elements T="BI" S="1">1</elements>
    <elements T="BI" S="2">2</elements>
    <elements T="BI" S="3">3</elements>
  </name>
  <name>
    <elements T="BM" S="1">10</elements>
    <elements T="BM" S="2">20</elements>
    <elements T="BM" S="3">30</elements>
  </name>
  <name>
    <elements T="XX" S="1">001</elements>
    <elements T="XX" S="2">002</elements>
  </name>
</root>

Sample Target Output #

Let's say we want to restructure each <name> into a <data-group> where each unique type (T) gets its own container with sequence items:

<root>
  <data-group>
    <type id="BI">
      <sequence number="1">1</sequence>
      <sequence number="2">2</sequence>
      <sequence number="3">3</sequence>
    </type>
  </data-group>
  <data-group>
    <type id="BM">
      <sequence number="1">10</sequence>
      <sequence number="2">20</sequence>
      <sequence number="3">30</sequence>
    </type>
  </data-group>
  <data-group>
    <type id="XX">
      <sequence number="1">001</sequence>
      <sequence number="2">002</sequence>
    </type>
  </data-group>
</root>

XSLT Code (2.0+) #

This uses XSLT 2.0's grouping features, which are widely supported in modern processors (like Saxon, Xalan, or built-in tools in Java/.NET):

<xsl:stylesheet version="2.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
  <!-- Identity template: copy all nodes/attributes by default -->
  <xsl:template match="@*|node()">
    <xsl:copy>
      <xsl:apply-templates select="@*|node()"/>
    </xsl:copy>
  </xsl:template>

  <!-- Transform <name> to <data-group> -->
  <xsl:template match="name">
    <data-group>
      <!-- Group <elements> by their T attribute -->
      <xsl:for-each-group select="elements" group-by="@T">
        <type id="{current-grouping-key()}">
          <!-- Process each element in the group -->
          <xsl:for-each select="current-group()">
            <sequence number="{@S}">
              <xsl:value-of select="."/>
            </sequence>
          </xsl:for-each>
        </type>
      </xsl:for-each-group>
    </data-group>
  </xsl:template>
</xsl:stylesheet>

Key Explanations #

• Identity Template: This is the foundation of most XSLT transformations—it copies everything from the input unless we override it with a specific template. That means any parts of your XML not mentioned in other templates (like the <root> element) will stay as-is.
• Grouping with xsl:for-each-group: This is the "advanced" part you mentioned. It groups all <elements> under a <name> by their T attribute. current-grouping-key() gives us the value of T for the current group (like "BI" or "BM").
• Dynamic Attribute Values: We use curly braces {} to insert attribute values directly from the input (e.g., id="{current-grouping-key()}" takes the T value and puts it into the id attribute of <type>).

Adapting to Your Exact Needs #

If your target structure is different, just modify the templates:

• If you want to keep <name> instead of <data-group>, remove the <data-group> wrapper and adjust the template match to just process the children.
• If you need a flat structure (no grouping by T), skip the xsl:for-each-group and just transform each <elements> directly to your desired element.
• If you need to filter certain T values, add a where clause to the xsl:for-each-group (e.g., group-by="@T" where @T != 'XX').

内容的提问来源于stack exchange，提问作者qubiter