Using xml-to-json. "Inner XML/XHTML" should go in a json string value field - json

Question

I'm using Saxon Home Edition to convert XML to JSON:
<?xml version="1.0" encoding="UTF-8"?>
<xsl:stylesheet version="3.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
<xsl:output method="text" encoding="UTF-8" />
<xsl:template match="/">
<xsl:variable name="xmljson">
<map xmlns="http://www.w3.org/2005/xpath-functions">
<string key="name">Some name</string>
<string key="description">A nice description</string>
</map>
</xsl:variable>
<xsl:value-of select="xml-to-json($xmljson)" />
</xsl:template>
</xsl:stylesheet>
Produces the desired output:
{"name":"Some name","description":"A nice description"}
The description field contains arbitrary complex xhtml. The template rule does not work with the following description field:
<string key="description">A <strong>nice</strong> description</string>
Error message:
xml-to-json: unknown element <strong>
Enclosing the description in a CDATA section does work:
<string key="description"><![CDATA[A <strong>nice</strong> description]]></string>
Desired output:
{"name":"Some name","description":"A <strong>nice<\/strong> description"}
Problem/Question
The content of the description field is the result of a transformation. So with and withou CDATA fails. This will not work:
<string key="description"><xsl:apply-template select="description" /></string>
Neither this:
<string key="description"><![CDATA[<xsl:apply-template select="description" />]]></string>

Answer 1

Use the serialize() function to produce escaped XML markup as text.
<xsl:variable name="options" as="element()">
<output:serialization-parameters xmlns:output="http://www.w3.org/2010/xslt-xquery-serialization">
<output:omit-xml-declaration value="yes"/>
</output:serialization-parameters>
</xsl:variable>
<xsl:variable name="xmljson" as="element()">
<map xmlns="http://www.w3.org/2005/xpath-functions">
<string key="name">Some name</string>
<string key="description"><xsl:value-of select="serialize(description, $options)"/></string>
</map>
</xsl:variable>

Answer 2

I came up with the following quick hack. Have to check this tomorrow again, kind of tired now...t
Edit: This is a horrible hack! The template rule does not treat special characters. Please read the comments below.
<?xml version="1.0" encoding="UTF-8"?>
<xsl:stylesheet version="3.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
<xsl:output method="text" encoding="UTF-8" />
<xsl:template match="/">
<xsl:variable name="doc">
<description>A <strong class="red">nice</strong> description</description>
</xsl:variable>
<xsl:variable name="xmljson">
<map xmlns="http://www.w3.org/2005/xpath-functions">
<string key="description"><xsl:apply-templates select="$doc/description/text()|$doc/description/*" /></string>
</map>
</xsl:variable>
<xsl:value-of select="xml-to-json($xmljson)" />
</xsl:template>
<xsl:template match="*">
<!-- opening tag -->
<xsl:text><</xsl:text>
<xsl:value-of select="name()"/>
<!-- attribute nodes -->
<xsl:apply-templates select="#*"/>
<xsl:text>></xsl:text>
<xsl:apply-templates/>
<!-- closing tag -->
<xsl:text><</xsl:text>
<xsl:text>/</xsl:text>
<xsl:value-of select="name()"/>
<xsl:text>></xsl:text>
</xsl:template>
<xsl:template match="#*">
<xsl:text> </xsl:text>
<xsl:value-of select="name()"/>
<xsl:text>="</xsl:text>
<xsl:value-of select="." />
<xsl:text>"</xsl:text>
</xsl:template>
</xsl:stylesheet>
Produces the desired output:
{"description":"A <strong class=\"red\">nice<\/strong> description"}