XML 和样式表 (v1.0)：从同一记录返回多个值的问题

Answer 1

看看这是否适合您：

XSLT 1.0

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

<xsl:template match="/PFA">
    <data>
        <record skip="true">
            <value name="Id" type="string"/>
            <value name="NameType" type="string"/>
            <value name="Suffix" type="string"/>
            <value name="CarName" type="string"/>
        </record>
        <xsl:for-each select="Records/Car">
            <xsl:variable name="id" select="@id" />
            <xsl:for-each select="NameDetails/Name">
                <xsl:variable name="nType" select="@NameType" />
                <xsl:for-each select="NameValue">
                    <record>
                        <value name="Id">
                            <xsl:value-of select="$id"/>
                        </value>
                        <value name="NameType">
                            <xsl:value-of select="$nType"/>
                        </value>
                        <value name="Suffix">
                            <xsl:value-of select="Suffix"/>
                        </value>
                        <value name="CarName">
                            <xsl:value-of select="CarName"/>
                        </value>
                    </record>       
                    </xsl:for-each>
            </xsl:for-each>
        </xsl:for-each>
    </data>
</xsl:template>

</xsl:stylesheet>