Question

For the below xml i want to render the unescaped text parts on browser from xslt e.g.:

<one>
&lt;text&gt;
</one>

I want the markup rendered on browser to be:

<text>

But when I use the apply templates which looks like below

<xsl:template match="text()" mode="literalHTML">
    <xsl:copy-of select=".">
    </xsl:copy-of>
</xsl:template>

The above XML is getting rendered as:

<text>

How can I modify this template so that it prints <text> on browser?

Best Regards,
Keshav

Answer 1

This can be achieved in XSLT 1.0 using quite tricky recursive processing.

Fortunately, one can use FXSL ( a library of XSLT templates) to solve the same task in just a few minutes:

<xsl:stylesheet version="1.0" 
xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
xmlns:f="http://fxsl.sf.net/"
xmlns:testmap="testmap"
exclude-result-prefixes="xsl f testmap">
   <xsl:import href="str-dvc-map.xsl"/>

   <testmap:testmap/>

   <xsl:output omit-xml-declaration="yes" indent="yes"/>

   <xsl:template match="/">
     <xsl:variable name="vTestMap" select="document('')/*/testmap:*[1]"/>
     <xsl:call-template name="str-map">
       <xsl:with-param name="pFun" select="$vTestMap"/>
       <xsl:with-param name="pStr" select="/*/text()"/>
     </xsl:call-template>
   </xsl:template>

    <xsl:template name="escape" mode="f:FXSL"
     match="*[namespace-uri() = 'testmap']">
      <xsl:param name="arg1"/>

      <xsl:choose>
       <xsl:when test="$arg1 = '&lt;'">&amp;lt;</xsl:when>
       <xsl:when test="$arg1 = '&gt;'">&amp;gt;</xsl:when>
       <xsl:otherwise><xsl:value-of select="$arg1"/></xsl:otherwise>
      </xsl:choose>
    </xsl:template>

</xsl:stylesheet>

when this transformation is applied on the following XML document:

<one>
&lt;text&gt;
</one>

the wanted result is produced:

<text>

and it displays in the browser as: <text>