用于 XML 到 XML 转换的高级 XSLT 转换

Question

我有一个如下所示的 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>
    <elements T="XX" S="3">003</elements>
</name>
<name>
    <elements T="XX" S="1">005</elements>
    <elements T="XX" S="2">007</elements>
    <elements T="XX" S="3">009</elements>
</name> </root>

我正在尝试使用以下 XSLT 转换将此 XML 转换为另一个 XML：

<xsl:for-each select="root">
<name_code_1>
    <xsl:for-each select="name">
        <xsl:for-each select="element[not(@T=../preceding-sibling::*/element/@T)]">
            <xsl:if test="@T = 'XX' and @S = '1'">
                <xsl:value-of select="substring(./text(),1,1000)"/>
            </xsl:if>
        </xsl:for-each>
    </xsl:for-each>
</name_code_1>
<name_code_2>
    <xsl:for-each select="name">
        <xsl:for-each select="element[not(@T=../preceding-sibling::*/element/@T)]">
            <xsl:if test="@T = 'XX' and @S = '2'">
                <xsl:value-of select="substring(./text(),1,1000)"/>
            </xsl:if>
        </xsl:for-each>
    </xsl:for-each>
</name_code_2>
<name_code_3>
    <xsl:for-each select="name">
        <xsl:for-each select="element[not(@T=../preceding-sibling::*/element/@T)]">
            <xsl:if test="@T = 'XX' and @S = '3'">
                <xsl:value-of select="substring(./text(),1,1000)"/>
            </xsl:if>
        </xsl:for-each>
    </xsl:for-each>
</name_code_3>
<name_code_4>
    <xsl:for-each select="name">
        <xsl:for-each select="element[not(@T=../preceding-sibling::*/element/@T)]">
            <xsl:if test="@T = 'XX' and @S = '1'">
                <xsl:value-of select="substring(./text(),1,1000)"/>
            </xsl:if>
        </xsl:for-each>
    </xsl:for-each>
</name_code_4>
<name_code_5>
    <xsl:for-each select="name">
        <xsl:for-each select="element[not(@T=../preceding-sibling::*/element/@T)]">
            <xsl:if test="@T = 'XX' and @S = '2'">
                <xsl:value-of select="substring(./text(),1,1000)"/>
            </xsl:if>
        </xsl:for-each>
    </xsl:for-each>
</name_code_5>
<name_code_6>
    <xsl:for-each select="name">
        <xsl:for-each select="element[not(@T=../preceding-sibling::*/element/@T)]">
            <xsl:if test="@T = 'XX' and @S = '3'">
                <xsl:value-of select="substring(./text(),1,1000)"/>
            </xsl:if>
        </xsl:for-each>
    </xsl:for-each>
</name_code_6> </xsl:for-each>

我想展平 T-attribute="XX" 并基于 S-attribute="1" 或 "2" 等的名称标签...将这些元素填充到 6 个不同的 XML 标签中，如下所示. 上面的逻辑不工作。有人可以通过提供一些有关如何解决此问题的建议来帮助我。

以下是所需的最终输出：

<some_tag>
<name_code_1>001</name_code_1>
<name_code_2>002</name_code_2>
<name_code_3>003</name_code_3>
<name_code_4>005</name_code_4>
<name_code_5>007</name_code_5>
<name_code_6>009</name_code_6> </some_tag>

谢谢 量子比特

Answer 1

一种方法是使用position() 函数从elements 节点的子集中获取节点索引：

<?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" omit-xml-declaration="yes" indent="yes"/>

  <xsl:template match="text()" />                              <!-- remove superfluous text() nodes -->

  <xsl:template match="/root">                                 <!-- main match and creation of <some_tag> element -->
    <some_tag>
      <xsl:apply-templates select="name/elements[@T='XX']" />  <!-- create subset of nodes with attribute @T='XX' -->
    </some_tag>
  </xsl:template>

  <xsl:template match="elements">                              <!-- match elements from subset list -->
    <xsl:element name="{concat('name_code_',position())}">     <!-- concat position to node-name -->
      <xsl:value-of select="." />                              <!-- copy value -->
    </xsl:element>
  </xsl:template>

</xsl:stylesheet>

输出为：

<some_tag>
    <name_code_1>001</name_code_1>
    <name_code_2>002</name_code_2>
    <name_code_3>003</name_code_3>
    <name_code_4>005</name_code_4>
    <name_code_5>007</name_code_5>
    <name_code_6>009</name_code_6>
</some_tag>

Answer 2

此方法使用 for-each-group，如果您使用 XSLT 2.0，这是最好的方法。

<?xml version="1.0" encoding="UTF-8"?>
<xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
    version="2.0">
    <xsl:output method="xml" indent="yes"/>
    <xsl:template match="root">
        <xsl:for-each-group select="name/elements" group-adjacent="if(@T='XX') then 'element' else 'cc'">
            <xsl:choose>
                <xsl:when test="current-grouping-key() = 'element'">
                    <some_tag>
                        <xsl:for-each select="current-group()">
                            <xsl:variable name="ss" select="position()"/>
                            <xsl:element name="{concat('name_code_', $ss)}">
                                <xsl:value-of select="."/>
                            </xsl:element>
                        </xsl:for-each>
                     </some_tag>
                </xsl:when>
                <xsl:otherwise></xsl:otherwise>
            </xsl:choose>
        </xsl:for-each-group>
    </xsl:template>
</xsl:stylesheet>

以下是上述 XSLT 生成的输出。

<?xml version="1.0" encoding="UTF-8"?>
<some_tag>
   <name_code_1>001</name_code_1>
   <name_code_2>002</name_code_2>
   <name_code_3>003</name_code_3>
   <name_code_4>005</name_code_4>
   <name_code_5>007</name_code_5>
   <name_code_6>009</name_code_6>
</some_tag>