XML Transformation using XSLT

Question

Hi,
I have a scenario where I need to perform an XML to XML transformation and I need to do this using XSLT mapping.
I am using XMLSpy.
Could somebody help me on this issue?Any weblogs that can help me on this?
regards,
Prashanth

Answer 1

Hi Prashanth,
You can do the xsl transformation in XMLSpy itself.First create your source xml file in XMLSpy by selecting file type as xml.After that create your xslt program file by selecting file type as xslt.
Then you can use the xslt program for transformation of source xml into target xml by choosing menu XSL/XQuery->XSL Transformation.
sample xsl code for an example:
source xml structure:
<?xml version="1.0" encoding="UTF-8"?>
<root>
     <header>
          <detail>
Detail1
</detail>
          <detail>
Detail2
</detail>
     </header>
     <header>
          <detail>
Detail3
</detail>
     </header>
</root>
xsl program:
<xsl:stylesheet version = '1.0'
     xmlns:xsl='http://www.w3.org/1999/XSL/Transform'>
<xsl:template match="/">
     <root>
     <xsl:for-each select="//header">
           <idoc>
            <xsl:for-each select="detail">
                <detail>
                      <xsl:value-of select="."/>
                   </detail>
                   </xsl:for-each>
               </idoc>
          </xsl:for-each>
     </root>
</xsl:template>
</xsl:stylesheet>
target xml structure:
<?xml version="1.0" encoding="UTF-8"?>
<root>
     <idoc>
          <detail>
Detail1
</detail>
          <detail>
Detail2
</detail>
     </idoc>
     <idoc>
          <detail>
Detail3
</detail>
     </idoc>
</root>
Hope this would help.
Rgds
Sudhakar.