如何刪除XML註釋沒有XSLT

Question

輸入文件中刪除其文本

<a> 
<b> 
<!--<docinfo:topiccodes><docinfo:topiccode>#PA#HKCRP#</docinfo:topiccode> <docinfo:topiccode>#PA#HKCRPM#</docinfo:topiccode><docinfo:topiccode>#PA#HKDIS#</docinfo:topiccode><docinfo:topiccode>#PA#HKDISM#</docinfo:topiccode><docinfo:topiccode>#PA#CODE#</docinfo:topiccode></docinfo:topiccodes>--> 
</b> 
</a> 

輸出expexted

<a> 
<b> 
<docinfo:topiccodes><docinfo:topiccode>#PA#HKCRP#</docinfo:topiccode> <docinfo:topiccode>#PA#HKCRPM#</docinfo:topiccode><docinfo:topiccode>#PA#HKDIS#</docinfo:topiccode><docinfo:topiccode>#PA#HKDISM#</docinfo:topiccode><docinfo:topiccode>#PA#CODE#</docinfo:topiccode></docinfo:topiccodes> 
</b> 
</a> 

XSLT書面

<xsl:template match="comment()"> 
<xsl:value-of select="."/> 
</xsl:template> 

Answer 1

試試這個：

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

    <xsl:strip-space elements="*"/> 
    <xsl:output indent="yes" omit-xml-declaration="yes"/> 

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

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

</xsl:stylesheet>