Custom Root Node in XSLT

Question

I want to create root node in xslt 1.0 in custom fashion

Expected

" < TESTROOT xmlns="http://www.example.org/TESTXMLSchema" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xsi:schemaLocation="http://www.example.org/TESTXMLSchema TESTEntry.xsd">

Actual

" < TESTROOT xsi:schemaLocation="http://www.example.org/TESTXMLSchema TESTEntry.xsd" xmlns="xmlns="http://www.example.org/TESTXMLSchema"" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance">

Thanks for your help in advance

Regards Rameshkumar singh

Dimitre Novatchev · Accepted Answer · 2012-10-16 15:13:00Z

3

As simple as this:

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

 <xsl:template match="/">
  <TESTROOT xmlns="http://www.example.org/TESTXMLSchema"
            xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" 
            xsi:schemaLocation="http://www.example.org/TESTXMLSchema TESTEntry.xsd">
    The results of your processing here ...
  </TESTROOT>
 </xsl:template>
</xsl:stylesheet>

When this transformation is applied on any XML document (not used), the wanted result is produced:

<TESTROOT xmlns="http://www.example.org/TESTXMLSchema" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xsi:schemaLocation="http://www.example.org/TESTXMLSchema TESTEntry.xsd">
    The results of your processing here ...
  </TESTROOT>

answered Oct 16, 2012 at 15:13

Dimitre Novatchev

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5 Comments

Dimitre Novatchev Over a year ago

@RameshSingh, What does it mean "not working" just copy and paste the transformation from the answer ( as I did when I verified that running the transformation produces the wanted result) and run the transformation -- on any source XML document -- then any compliant XSLT processor produces the result that is provided (again copied from the actual result of running the transformation and pasted into this answer) in this answer. I always test my code and verify that it is actually producing the wanted result.

Ramesh Singh Over a year ago

hi dimitre, i use xslt transformation through biztalk pipeline component . when it parses through parser automatically it is back to actual results .

Dimitre Novatchev Over a year ago

@RameshSingh, BizTalk is not an XSLT processor. You must read the documentation and see what restrictions/conventions for XSLT transformation do apply. I am not a BizTalk specialist, but know my XSLT stuff. I have verified that all of the 10 different XSLT processors I am working with produce the same wantes, correct result -- MSXML 3,4,6, Saxon 6.5.4, Saxon 9.1.05, AltovaXML (XML-SPY) -- both for XSLT 1.0 and XSLT 2.0, .NET XslCompiledTransform, .NET XslTransform, XQSharp (XMLPrime).

Ramesh Singh Over a year ago

I tested using visual studio iam getting the Actual results

Dimitre Novatchev Over a year ago

@RameshSingh, I am glad my answer was useful to you. Could you, please, accept it (by clicking the check-mark next to the answer)?

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