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>
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