How to convert JSON to XML using XSLT? - json

Question

I want to convert JSON to XML using XSLT. But not able to achieve the expected output.
Below is the JSON request:
{
"Store": [
{
"Book": "Cartoons",
"ID": "ABC"
}
]
}
The XSLT which I tried:
<xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
xmlns:xs="http://www.w3.org/2001/XMLSchema"
xmlns:math="http://www.w3.org/2005/xpath-functions/math"
xmlns:xd="http://www.oxygenxml.com/ns/doc/xsl"
xmlns:emp="http://www.semanticalllc.com/ns/employees#"
xmlns:h="http://www.w3.org/1999/xhtml"
xmlns:fn="http://www.w3.org/2005/xpath-functions"
xmlns:j="http://www.w3.org/2005/xpath-functions"
exclude-result-prefixes="xs math xd h emp"
version="3.0"
expand-text="yes">
<xsl:template match="/">
<xsl:copy>
<xsl:apply-templates select="json-to-xml(.)/*"/>
</xsl:copy>
</xsl:template>
<xsl:template match="*[#key]" xpath-default-namespace="http://www.w3.org/2005/xpath-functions">
<xsl:element name="{#key}">
<xsl:apply-templates/>
</xsl:element>
</xsl:template>
</xsl:stylesheet>
But I'm getting empty Response.

Answer 1

You need to pass the JSON as a parameter or read it from a file, the input to your XSLT is either XML or you can start with a named template:
<?xml version="1.0" encoding="UTF-8"?>
<xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
xmlns:xs="http://www.w3.org/2001/XMLSchema"
exclude-result-prefixes="#all"
version="3.0">
<xsl:param name="json" as="xs:string" expand-text="no">{
"Store": [
{
"Book": "Cartoons",
"ID": "ABC"
}
]
}</xsl:param>
<xsl:output indent="yes"/>
<xsl:template match="/" name="xsl:initial-template">
<xsl:sequence select="json-to-xml($json)"/>
</xsl:template>
</xsl:stylesheet>
https://xsltfiddle.liberty-development.net/93wniTJ
The result of the function for your sample is
<map xmlns="http://www.w3.org/2005/xpath-functions">
<array key="Store">
<map>
<string key="Book">Cartoons</string>
<string key="ID">ABC</string>
</map>
</array>
</map>
but of course you can run it through further templates to transform it to a different format.

Answer 2

The pattern match="/" matches a document node (the root of an XML tree). It won't match your input if the input is JSON.
XSLT 3.0 isn't actually that good at processing JSON using template rules: it can be done, but it isn't very convenient. It's usually more convenient to use functions. You can supply the JSON input as the value of an xsl:param and process it in code from your xsl:initial-template; or if you're feeling more ambitious you could actually initiate the XSLT processing by invoking a public xsl:function that takes the JSON input as a parameter.
The traditional match="/" entry to a stylesheet only works for the traditional use of XSLT to process an XML input document.