How do I rename a node?

What language are you going to perform the conversion in?  I don't think you can do this in XSL, but who knows...

hi Shaun,
Since node Name is a ready only property, the solution i have suggested is a bit of a round about way.

The function requires 3 parameters to make it generic.
viz. oldName,newName and parentName

What u do here is remove the <oldName> node after keeping a pointer to all its children.

Create a new node with the name <newName> and append it
to parent Node. Then to this node append all the childeren to which we had kept pointers earlier.


function changeNodeName(oldName,newName,parentName) {
          xmldom = new ActiveXObject("MICROSOFT.XMLDOM")
          xmldom.load(zvon1.XMLDocument);
         
          t_ptr = xmldom.selectSingleNode("//" + oldName).childNodes    
          xmldom.selectSingleNode("//" + parentName).removeChild(xmldom.selectSingleNode("//" + oldName))
         
          newNode = xmldom.createNode(1,newName,"")          
          xmldom.selectSingleNode("//" + parentName).appendChild(newNode)
         
          for(i=0;i<t_ptr.length;++i) {
               xmldom.selectSingleNode("//" + parentName).childNodes(0).appendChild(t_ptr(i))
          }
          alert(xmldom.xml)
     }

I fell this code must work and hope it will be useful to.

rgds
xav

Here is your XSLT stylesheet:

<?xml version="1.0"?>
<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform" xsl:space="preserve">

<xsl:template match="a">
     <first><xsl:apply-templates/></first>
</xsl:template>

<xsl:template match="b">
     <second><xsl:apply-templates/></second>
</xsl:template>

<xsl:template match="/ | @* | node()">
     <xsl:copy><xsl:apply-templates select="@* | node()"/></xsl:copy>
</xsl:template>

</xsl:stylesheet>

hi Shaun,
Since node Name is a ready only property, the solution i have suggested is a bit of a round about way.

The function requires 3 parameters to make it generic.
viz. oldName,newName and parentName

What u do here is remove the <oldName> node after keeping a pointer to all its children.

Create a new node with the name <newName> and append it
to parent Node. Then to this node append all the childeren to which we had kept pointers earlier.


function changeNodeName(oldName,newName,parentName) {
          xmldom = new ActiveXObject("MICROSOFT.XMLDOM")
          xmldom.load(zvon1.XMLDocument);
         
          t_ptr = xmldom.selectSingleNode("//" + oldName).childNodes    
          xmldom.selectSingleNode("//" + parentName).removeChild(xmldom.selectSingleNode("//" + oldName))
         
          newNode = xmldom.createNode(1,newName,"")          
          xmldom.selectSingleNode("//" + parentName).appendChild(newNode)
         
          for(i=0;i<t_ptr.length;++i) {
               xmldom.selectSingleNode("//" + parentName).childNodes(0).appendChild(t_ptr(i))
          }
          alert(xmldom.xml)
     }

I fell this code must work and hope it will be useful to.

rgds
xav

sorry guys, i first gave the answer y'day.
came today and refreshed the page and lo,
my answer gets posted again. Me real sorry for it, but to some extent i feel it is the fault of the way this site ( i am only reffering to this functionality)  is build.
When i submit it should go to a page wihich inserts into the db my commenst/answers, redirect to another page which is the view page. Now if i refresh, only the view generating page is refreshed, so  repeat d/b entry does not happen.
Hope the suggestion is take in the right spirit.
cheers
xav

xaviergeorge - that is why there is a Reload Question option at the top of the page - as I remember it will not let you post the same message multiple times within a certain time frame

chabaud - I was working on someting similar - I'll check yours out - my XSLT is awful :( - will the atributes be preserved also?

chabaud - it works - with some mods as it didn't preserve the attributes of the node so I used xsl:copy-of eg

<xsl:template match="a">
    <first><xsl:copy-of select="@*"/><xsl:apply-templates/></first>
</xsl:template>

just one extra (if you don't mind)

if I had

<a NLSname="A" ...>
  <b NLSname="B" ...>
  </b>
</a>

how do I change it to

<X NLSname="x" ...>
  <Y NLSname="y" ...>
  </Y>
</X>

eg <a> always has the NLSName == A and I wish to change it to <X> with NLSName = x etc

I actually did it this way

<xsl:template match="a">
    <first>
          <xsl:copy-of select="@*"/>
          <xsl:attribute name="Name">x</xsl:attribute>
          <xsl:apply-templates/>
     </first>
</xsl:template>

but the effect is the same

thanks - could you possibly comment on some of the lines and what they do  eg what @* means etc

For your information,

@   attribute prefix
@*  all attributes matches

For more example and documentation (from Microsoft):

Patterns:
---------
http://msdn.microsoft.com/library/default.asp?url=/library/en-us/xmlsdk30/htm/xmconpatterns.asp

XSL developers guide:
---------------------
http://msdn.microsoft.com/library/default.asp?url=/library/en-us/xmlsdk30/htm/xmconxsldevelopersguide.asp

XSLT reference:
---------------
http://msdn.microsoft.com/library/default.asp?url=/library/en-us/xmlsdk30/htm/xmrefxsltreference.asp

thanks again - look out for more XSLT related questions from me - I am sure ther will be some