Example XSLT to Convert BookQuery XML Response to JSON Response

Use the following XSLT to convert XML message format to JSON message format: 

<xsl:stylesheet version="1.0"  xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
xmlns:ns0="http://www.example.com/xsd/books"
xmlns:fn="http://www.tibco.com/asg/functions/json"
xmlns:json="http://www.tibco.com/asg/content-types/json"
xmlns:SOAP-ENV="http://schemas.xmlsoap.org/soap/envelope/"
exclude-result-prefixes="xsl fn json ns0 SOAP-ENV">
<xsl:output method="xml" version="1.0" encoding="UTF-8" indent="yes" omit-xml-declaration="yes"/>
<xsl:variable name="cnResponseHref">
<xsl:value-of select="/transformation/cnResponse/@href"/>
</xsl:variable>
<xsl:variable name="cnResponse">
<xsl:copy-of select="document($cnResponseHref)/*"/>
</xsl:variable>
<xsl:variable name="JsonXmlResponse">
<json:json_xml>
<json:dict>
<json:dict key="BookStore">
<json:list key="Book">
<xsl:for-each select="$cnResponse/SOAP-ENV:Envelope/SOAP-ENV:Body/ns0:BookStore/ns0:Book">
<json:dict key="Book">
<json:text key="Title"><xsl:value-of select="ns0:Title"/></json:text>
<json:text key="ISBN"><xsl:value-of select="ns0:ISBN"/></json:text>
<json:text key="Author"><xsl:value-of select="ns0:Author"/></json:text>
<json:text key="Publisher"><xsl:value-of select="ns0:Publisher"/></json:text>
</json:dict>
</xsl:for-each>
</json:list>
</json:dict>
</json:dict>
</json:json_xml>
</xsl:variable>
<xsl:template match="/">
<xsl:value-of select="fn:render($JsonXmlResponse//json:json_xml/json:dict)"/>
</xsl:template>
</xsl:stylesheet>