[XSL-LIST Mailing List Archive Home] [By Thread] [By Date]

Re: [xsl] element name as attribute value

---

---

Hi,

You can just start with the recursive copy template and add specific rules to handle the specific processing you want in different contexts. For instance the example below changes X in A and changes Y in <B attribute="Y">

<?xml version="1.0" encoding="UTF-8"?>
<xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform" version="2.0">
<xsl:template match="node() | @*">
<xsl:copy>
<xsl:apply-templates select="node() | @*"/>
</xsl:copy>
</xsl:template>
<xsl:template match="X">
<A>
<xsl:apply-templates select="node() | @*"/>
</A>
</xsl:template>
<xsl:template match="Y">
<B attribute="Y">
<xsl:apply-templates select="node() | @*"/>
</B>
</xsl:template>
</xsl:stylesheet>

Best Regards,
George
---------------------------------------------------------------------
George Cristian Bina
<oXygen/> XML Editor, Schema Editor and XSLT Editor/Debugger
http://www.oxygenxml.com

San wrote:

Hi all,

Hope you can help me with this problem (seems easy but
somehow i can't figure how to do it :(( )
I have this xml file

<X>
  <Y/>
</X>

I would like to have the output (element Y as
attribute value from other element e.g. B)

<A>
  <B attribute="Y">
</A>

I tried with <xsl:copy-of select="child::node()"/> it worked but only in the element part, can't put it
as attribute value.
It become like

<A>
  <Y/>
</A>

Really appreciate when someone can help this desperate
person.

Many thanks before
san

____________________________________________________________________________________
Need Mail bonding?
Go to the Yahoo! Mail Q&A for great tips from Yahoo! Answers users.
http://answers.yahoo.com/dir/?link=list&sid=396546091

---