我有一个输入XML文档：

输入XML

<?xml version="1.0" encoding="UTF-8"?>
<allNames id="ID_0" b:type="a:UnstructuredName">
    <typeName>KnownBy</typeName>
    <startDate>2001-01-01-05:00</startDate>
    <fullName>ABCD 004 COMPANY INC</fullName>
</allNames>

我需要应用XSLT将其转换为

输出XML

<?xml version="1.0" encoding="UTF-8"?>
<allNames  b:type="a:UnstructuredName"  id="ID_0">
    <typeName>KnownBy</typeName>
    <startDate>2001-01-01-05:00</startDate>
    <fullName>ABCD 004 COMPANY INC</fullName>
</allNames>

唯一的变化是allNames元素中的属性排序更改.我查了另一篇文章并编写了命令属性的XSLT,但我不知道如何让整个过程发挥作用.

XSLT

<?xml version="1.0" encoding="UTF-8"?>
<xsl:stylesheet version="2.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform" xmlns:fo="http://www.w3.org/1999/XSL/Format" xmlns:xs="http://www.w3.org/2001/XMLSchema" xmlns:fn="http://www.w3.org/2005/xpath-functions">
    <xsl:output method="xml" indent="yes"/>
    <xsl:variable name="attributes" select="document('mytest.xml')//attribute"/>
    <xsl:template match="@* | node()">
        <xsl:copy>
            <xsl:apply-templates select="@* | node()"/>
        </xsl:copy>
    </xsl:template>
    <xsl:template match="*">
        <xsl:variable name="self" select="."/>
        <xsl:for-each select="$attributes">
            <xsl:apply-templates select="$self/@*[name()=current()]"/>
        </xsl:for-each>   
     </xsl:template>
</xsl:stylesheet>

mytest.xml

<?xml version="1.0" encoding="UTF-8"?>
    <attributes>
        <attribute>b:type</attribute>
        <attribute>id</attribute>
    </attributes>