"sg" <sg@d...> wrote in message 
news:AA136C57-0D8F-45C3-BCEA-947094A2B31E@m......
> hi,
>
> I have this xml:
> <root>
>    <book>
>        <name>book1</name
>    </book>
>    <book>
>        <name>book1</name
>    </book>
>    .
>    .
>    .
> </root>
>
> and I'm tring to transfer it to this:
> <root>
>    <book1>
>        <name>book1</name
>    </book>
>    <book2>
>        <name>book1</name
>    </book>
>    .
>    .
>    <bookn>
>        <name>book1</name
>    </book>
>
> How can I do this using XSL?

I wholeheartedly recommend against this, it's very difficult to process this 
type of markup, you can't easily sort or select all the book(n) elements.
This would be the stylesheet:

<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
<xsl:template match="/root">
<xsl:copy>
<xsl:apply-templates select="book"/>
</xsl:copy>
</xsl:template>
<xsl:template match="book">
<xsl:element name="{concat('book', position())}">
<xsl:copy-of select="name"/>
</xsl:element>
</xsl:template>
</xsl:stylesheet>

I'd prefer:

<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
<xsl:template match="/root">
<xsl:copy>
<xsl:apply-templates select="book"/>
</xsl:copy>
</xsl:template>
<xsl:template match="book">
<book position="{position()}">
<xsl:copy-of select="name"/>
</book>
</xsl:template>
</xsl:stylesheet>

Which would give:

<root>
  <book position="1">
    <name>book1</name>
  </book>
  <book position="2">
    <name>book1</name>
  </book>
...
</root>

-- 

Joe (MVP - XML)