Traverse json document using xslt

Question

I'm trying to fetch field inside json data

Input json

{
    "data": {
        "file": [{
                "id": "0001",
                "name": "harsha"
            },
            {
                "id": "0002",
                "name": "manohar"
            }
        ]
    }
}

XSLT style sheet

<xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
    xmlns:xs="http://www.w3.org/2001/XMLSchema" version="3.0"
    xmlns="http://www.w3.org/2005/xpath-functions" xpath-default-namespace="http://www.w3.org/2005/xpath-functions" expand-text="yes">
    <xsl:param name="input"/>
    <xsl:output method="text"/>
    
    <xsl:template name="xsl:initial-template">
        <xsl:variable name="input-as-xml" select="json-to-xml($input)"/>
        <xsl:variable name="transformed-xml" as="element(map)">
            
            <xsl:for-each select="$input-as-xml/map/array[@key='file']">
                <map>
                    <string key="date">
                    <xsl:value-of select="../string[@key='name']"/>
                </string>
                </map>
            </xsl:for-each>
           
        </xsl:variable>
        <xsl:value-of select="xml-to-json($transformed-xml)"/>
    </xsl:template>
</xsl:stylesheet>

Here I'm trying to fetch name field but not able to fetch I'm getting empty data

{"date":""}

Any suggestions would be helpful...

Start by looking at what the XML produced by json-to-xml($input) looks like. Then adjust your processing accordingly. — michael.hor257k
– michael.hor257k, Commented Jun 7, 2021 at 9:00
I don't see how you would get a map at all, you use <xsl:for-each select="$input-as-xml/map/array[@key='file']"> but that doesn't select anything in your input. And which is the resulting XML and/or JSON you want to produce? You have two name values so building a map with two properties of the same name is not going to work. — Martin Honnen
– Martin Honnen, Commented Jun 7, 2021 at 10:58

Michael Kay · Accepted Answer · 2021-06-07 11:00:42Z

1

The ".." looks wrong to me - why would you want the parent of the array element?

I would do this without conversion to/from XML. Something along the lines:

   <xsl:variable name="input-as-map" select="parse-json($input)" as="map(*)"/>
   <xsl:variable name="transformed-map" as="map(*)*">    
        <xsl:for-each select="$input-as-map?data?file?*">
             <xsl:map key="'date'">
                    <xsl:value-of select="?name"/>
             </xsl:map>
        </xsl:for-each>           
   </xsl:variable>
   <xsl:value-of select="serialize($transformed-map, map{'method':'json'}"/>

(untested, because I'm not sure what output you want).

answered Jun 7, 2021 at 11:00

Michael Kay

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Martin Honnen · Accepted Answer · 2021-06-07 11:19:35Z

0

It is also not clear to me what you want, perhaps

   <xsl:for-each select="$input-as-xml/map/map[@key = 'data']/array[@key='file']">
        <map>
            <string key="date">{.//string[@key='name']}</string>
        </map>
    </xsl:for-each>

gives you a value you can work with.

answered Jun 7, 2021 at 11:19

Martin Honnen

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