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Hello,

How can I write the XSLT to transform:

<Root>
<Element attribute="value"/>
<Element2>
<Element3/>
</Element2>
</Root>

into:

<Root>
<Element attribute="value"/>
<Element2>
<Element3/>
</Element2>
<New element>
....
</New element>
</Root>

The output is the same as the input with a new element at the end.

Thanks for your help.
Olivier

wrote:

> How can I write the XSLT to transform:
>
> <Root>
> <Element attribute="value"/>
> <Element2>
> <Element3/>
> </Element2>
> </Root>
>
> into:
>
> <Root>
> <Element attribute="value"/>
> <Element2>
> <Element3/>
> </Element2>
> <New element>
> ....
> </New element>
> </Root>
>
> The output is the same as the input with a new element at the end.

Start with the identity transformation template
<xsl:template match="@* | node()">
<xsl:copy>
<xsl:apply-templates select="@* | node()"/>
</xsl:copy>
</xsl:template>
then add
<xsl:template match="Root">
<xsl:copy>
<xsl:apply-templates select="@* | node()"/>
<New-Element>
....
</New-Element>
</xsl:copy>
</xsl:template>

--

Martin Honnen
http://JavaScript.FAQTs.com/

Martin,

Thanks very much, it works !

Olivier

On Jan 26, 4:51 pm, Martin Honnen <mahotr...@yahoo.de> wrote:
> Start with the identity transformation template
> <xsl:template match="@* | node()">
> <xsl:copy>
> <xsl:apply-templates select="@* | node()"/>
> </xsl:copy>
> </xsl:template>
> then add
> <xsl:template match="Root">
> <xsl:copy>
> <xsl:apply-templates select="@* | node()"/>
> <New-Element>
> ....
> </New-Element>
> </xsl:copy>
> </xsl:template>
>
> --
>
> Martin Honnen
> http://JavaScript.FAQTs.com/