如何使用 XSLT 注释掉特定元素

HOW TO COMMENT OUT PARTICULAR ELEMENT USING XSLT

如何仅使用 XSLT 取消对特定元素 'navref' 的注释,有人可以研究一下并提出建议吗?

XML 输入:

<?xml version="1.0" encoding="UTF-8"?>
<topicgroup collection-type="sequence">
<!--Handling topicref with navref-->
<topicref collection-type="sequence" id="107746"
    href="../../../../../../Section_General_Notes/107746.dita">
    <topicmeta>
        <navtitle/>
        <critdates>
            <created date="09-Apr-2009" golive="01-Jul-2009" expiry=""/>
        </critdates>
        <data name="section_num" value="55"/>
    </topicmeta>
</topicref>
<!--<navref mapref="../../../../../../Section_General_Notes/107746.ditamap"/>-->
</topicgroup>

预期输出:

<?xml version="1.0" encoding="UTF-8"?>
<topicgroup collection-type="sequence">
<!--Handling topicref with navref-->
<topicref collection-type="sequence" id="107746"
    href="../../../../../../FASB_Section_General_Notes/107746.dita">
    <topicmeta>
        <navtitle/>
        <critdates>
            <created date="09-Apr-2009" golive="01-Jul-2009" expiry=""/>
        </critdates>
        <data name="section_num" value="55"/>
    </topicmeta>
</topicref>
<navref mapref="../../../../../../Section_General_Notes/107746.ditamap"/>
</topicgroup>

XSLT:

<?xml version="1.0" encoding="UTF-8"?>
<xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
xmlns:xs="http://www.w3.org/2001/XMLSchema"
exclude-result-prefixes="xs"
version="2.0">

<xsl:template match="node() | @*">
    <xsl:copy>
        <xsl:apply-templates select="node() | @*"/>
    </xsl:copy>
</xsl:template>

<xsl:template match="comment()" priority="1">
    <xsl:value-of select="." disable-output-escaping="yes"/>
</xsl:template>

</xsl:stylesheet>

参考URL:https://xsltfiddle.liberty-development.net/6qtiBni/1

而不是:

<xsl:template match="comment()" priority="1">
    <xsl:value-of select="." disable-output-escaping="yes"/>
</xsl:template>

尝试:

<xsl:template match="comment()[starts-with(., '&lt;navref')]" >
    <xsl:value-of select="." disable-output-escaping="yes"/>
</xsl:template>