Ответ 1
Это преобразование:
<xsl:stylesheet version="1.0"
xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
xmlns:xs="http://www.w3.org/2001/XMLSchema" >
<xsl:output omit-xml-declaration="yes" indent="yes"/>
<xsl:strip-space elements="*"/>
<xsl:param name="ptopElementName" select="'RootElement'"/>
<xsl:variable name="vTop" select=
"/*/xs:element[@name=$ptopElementName]"/>
<xsl:variable name="vNames"
select="$vTop/descendant-or-self::*/@name"/>
<xsl:variable name="vRefs"
select="$vTop/descendant-or-self::*/@ref"/>
<xsl:variable name="vTypes"
select="$vTop/descendant-or-self::*/@type"/>
<xsl:template match="node()|@*" name="identity">
<xsl:copy>
<xsl:apply-templates select="node()|@*"/>
</xsl:copy>
</xsl:template>
<xsl:template match="xs:element">
<xsl:if test=
"@name=$vNames
or
@name=$vRefs
or
ancestor-or-self::*[@name=$ptopElementName]">
<xsl:call-template name="identity"/>
</xsl:if>
</xsl:template>
<xsl:template match="xs:complexType|xs:simpleType">
<xsl:if test=
"@name=$vTypes
or
ancestor-or-self::*[@name=$ptopElementName]">
<xsl:call-template name="identity"/>
</xsl:if>
</xsl:template>
</xsl:stylesheet>
при применении к предоставленному XML-документу:
<xs:schema xmlns:xs="http://www.w3.org/2001/XMLSchema" elementFormDefault="qualified" attributeFormDefault="unqualified">
<xs:element name="RootElement">
<xs:complexType>
<xs:sequence>
<xs:element ref="ChildElement"/>
</xs:sequence>
</xs:complexType></xs:element>
<xs:element name="ChildElement"/>
<xs:element name="UnusedElement"/>
</xs:schema>
создает желаемый результат:
<xs:schema xmlns:xs="http://www.w3.org/2001/XMLSchema" elementFormDefault="qualified" attributeFormDefault="unqualified">
<xs:element name="RootElement">
<xs:complexType>
<xs:sequence>
<xs:element ref="ChildElement"/>
</xs:sequence>
</xs:complexType>
</xs:element>
<xs:element name="ChildElement"/>
</xs:schema>