Need XSLT transformation to convert Input XML to Output XML

Question

HI,

I am looking to convert input XML to OUTput XML using XSLT.

Input XML is as below

<ALLFields id="0001">
  <field name="ComputerName">ABC</field>
  <field name="ComputerType">Windows</field>
  <field name="DatabaseName" />
  <field name="CPULevel">10</field>
</ALLFields>

OUTPUT XML what i need is

<entry id="0001">
<ComputerName>ABC</ComputerName>
<ComputerType>Windows</ComputerType>
<DatabaseName />
<CPULevel>10</CPULevel>
</entry>

Could you format your question so that the XML schema can be read and provide what you have already tried — detaylor
– detaylor, Commented Apr 7, 2011 at 11:57

Dimitre Novatchev · Accepted Answer · 2011-04-07 13:06:50Z

3

This transformation:

<xsl:stylesheet version="1.0"
 xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
 <xsl:output omit-xml-declaration="yes" indent="yes"/>
 <xsl:strip-space elements="*"/>

 <xsl:template match="ALLFields">
     <entry id="{@id}">
       <xsl:apply-templates select="node()"/>
     </entry>
 </xsl:template>

 <xsl:template match="field">
  <xsl:element name="{@name}">
   <xsl:apply-templates/>
  </xsl:element>
 </xsl:template>
</xsl:stylesheet>

when applied on the provided XML document (corrected to be well-formed):

<ALLFields id="0001">
    <field name="ComputerName">ABC</field>
    <field name="ComputerType">Windows</field>
    <field name="DatabaseName" />
    <field name="CPULevel">10</field>
</ALLFields>

produces the wanted, correct result:

<entry id="0001">
   <ComputerName>ABC</ComputerName>
   <ComputerType>Windows</ComputerType>
   <DatabaseName/>
   <CPULevel>10</CPULevel>
</entry>

answered Apr 7, 2011 at 13:06

Dimitre Novatchev

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1 Comment

Chazt3n Over a year ago

Using this code:

var myXslTrans = new XslCompiledTransform();  myXslTrans.Load("Files/NaughtyXSLT_Transform.xsl"); myXslTrans.Transform("Files/NaughtyXSLT_InputXML.xml","NaughtyXSLT_OutputXML.xml");

Does not work, jeez I cannot edit this comment for some reason!

khachik · Accepted Answer · 2011-04-07 13:50:31Z

0

<xsl:template match="AllFields">
    <entry id="{@id}">
        <xsl:for-each select="field">
            <xsl:element name="{@name}">
                <xsl:value-of select="text()"/>
            </xsl:element>
        </xsl:for-each>
    </entry>
</xsl:template>

Not tested though.

edited Apr 7, 2011 at 13:50

answered Apr 7, 2011 at 12:08

khachik

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2 Comments

Dimitre Novatchev Over a year ago

There is no element named entry in the provided source XML document.

khachik Over a year ago

It was <AllFields> ... </entry> so I chose entry. Thanks, updated.

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