xslt 1.0-使用xslt 1.0如何使xslt适用于多个文档元素

Question

这是我的XML-

<?xml version="1.0" encoding="UTF-8"?>
        <update-all-attributes>
          <document name="http://blah">
            <price>111 USD</price>
            <color>red</color>
            <size>comfort</size>
            <cost>00012-40</cost>
            <shipping>US::Ground:0.00</shipping>
            <cost>00012-40</cost>
            <price>111 USD</price>
            <color>red</color>
            <size>comfort</size>
            <cost>00012-40</cost>
            <shipping>US::Ground:0.00</shipping>
            <cost>00012-40</cost>
            <price>111 USD</price>
            <color>red</color>
            <size>comfort</size>
            <shipping>US::Ground:0.00</shipping>
            <price>111 USD</price>
            <color>red</color>
            <size>comfort</size>
          </document>


<document name="http://blahblah">
    <price>110 USD</price>
    <color>blue</color>
    <size>super</size>
    <cost>00012-41</cost>
    <shipping>US::Ground:0.01</shipping>
    <cost>00012-41</cost>
    <price>110 USD</price>
    <color>blue</color>
    <size>super</size>
    <shipping>US::Ground:0.01</shipping>
    <cost>00012-41</cost>
    <price>110 USD</price>
    <color>blue</color>
    <size>super</size>
    <shipping>US::Ground:0.01</shipping>
    <price>110 USD</price>
    <color>blue</color>
    <size>super</size>
  </document>
</update-all-attributes>

我希望我的最终XML为

<?xml version="1.0" encoding="UTF-8"?>
                <document name="http://blah">
                  <price>111 USD</price>
                  <color>red</color>
                  <size>comfort</size>
                  <cost>00012-40,00012-40,00012-40,00012-40</cost>
                  <shipping>US::Ground:0.00,US::Ground:0.00,US::Ground:0.00</shipping>
                </document>
            <document name="http://blahblah">
                  <price>110 USD</price>
                  <color>blue</color>
                  <size>super</size>
                  <cost>00012-41,00012-41,00012-41,00012-41</cost>
                  <shipping>US::Ground:0.01,US::Ground:0.01,US::Ground:0.01</shipping>
                </document>

这是使用的XSLT（感谢Thomas W）-

<?xml version="1.0" encoding="UTF-8"?>
<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="element-by-name" match="document/*" use="local-name()"/>


  <!-- Copy everything, unless other templates say otherwise -->
  <xsl:template match="*|@*">
    <xsl:copy>
      <xsl:apply-templates/>
    </xsl:copy>
  </xsl:template>


  <!-- Skip elements if there are elements of the same name earlier in the document. 
       Priority is set to 1 so that this template overrides other templates
       and we don't accidentally still get duplicates. -->
  <xsl:template match="*[count(. | key('element-by-name', local-name())[1]) > 1]" priority="1"/>


  <!-- This template is for elements we want to be multi-value -->
  <xsl:template match="shipping | cost">
    <xsl:copy>
      <!-- We step through all elements of the same name -->
      <xsl:for-each select="key('element-by-name', local-name())">
        <xsl:value-of select="."/>
        <xsl:if test="position() != last()">,</xsl:if>
      </xsl:for-each>
    </xsl:copy>
  </xsl:template>

</xsl:stylesheet>

这仅对一个文档元素有效，我有多个文档元素，那么如何对所有文档元素起作用？

Answer 1

您可以通过更改键来做到这一点，以使其同时使用元素名称和其document父级的唯一ID作为查找值。

<?xml version="1.0" encoding="UTF-8"?>
<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="element-by-name-and-parent" match="document/*" 
           use="concat(local-name(), generate-id(..))"/>

  <!-- Copy everything, unless other templates say otherwise -->
  <xsl:template match="*|@*">
    <xsl:copy>
      <xsl:copy-of select="@*"/>
      <xsl:apply-templates/>
    </xsl:copy>
  </xsl:template>

  <!-- Skip siblings of the same name within the same parent element; only use first occurrence -->
  <xsl:template match="*[count(. | key('element-by-name-and-parent', 
               concat(local-name(), generate-id(..)))[1]) > 1]"
        priority="1"/>

  <!-- This template is for elements we want to be multi-value -->
  <xsl:template match="shipping | cost">
    <xsl:copy>
      <!-- Step through all elements of the same name that belong to the same parent -->
      <xsl:for-each select="key('element-by-name-and-parent', 
                concat(local-name(), generate-id(..)))">
       <xsl:value-of select="."/>
        <xsl:if test="position() != last()">,</xsl:if>
      </xsl:for-each>
    </xsl:copy>
  </xsl:template>
</xsl:stylesheet>

输出（根元素与输入XML中的根元素相同；您在问题中要求的输出没有根元素）：

<?xml version="1.0" encoding="UTF-8"?>
<update-all-attributes>
   <document name="http://blah">
      <price>111 USD</price>
      <color>red</color>
      <size>comfort</size>
      <cost>00012-40,00012-40,00012-40,00012-40</cost>
      <shipping>US::Ground:0.00,US::Ground:0.00,US::Ground:0.00</shipping>
   </document>
   <document name="http://blahblah">
      <price>110 USD</price>
      <color>blue</color>
      <size>super</size>
      <cost>00012-41,00012-41,00012-41</cost>
      <shipping>US::Ground:0.01,US::Ground:0.01,US::Ground:0.01</shipping>
   </document>
</update-all-attributes>