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-xmldeclaration=" 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>