On 12 Jan., 12:42, Martin Honnen <mahotr...@yahoo.de> wrote:
> Michael Skovfoged wrote:
> > I want a node like this:
>
> > <RmtInf><Ustrd>Invoices 123 and 321</Ustrd></RmtInf>
>
> > to be output like this:
>
> > SOMEINFO =A0 =A0<RmtInf><Ustrd>Invoices 123 a
> > OTHERINFO =A0 nd 321</Ustrd></RmtInf>
>
> =A0 =A0<xsl:template match=3D"RmtInf">
> =A0 =A0 =A0<xsl:text>SOMEINFO =A0 =A0</xsl:text>
> =A0 =A0 =A0<xsl:copy>
> =A0 =A0 =A0 =A0<xsl:apply-templates/>
> =A0 =A0 =A0</xsl:copy>
> =A0 =A0</xsl:template>
>
> =A0 =A0<xsl:template match=3D"Ustrd">
> =A0 =A0 =A0<xsl:copy>
> =A0 =A0 =A0 =A0<xsl:value-of select=3D"concat(substring(., 1, 14), '&#10;=
OTHERINFO
> =A0 =A0', substring(., 15))"/>
> =A0 =A0 =A0</xsl:copy>
> =A0 =A0</xsl:template>
>
> > I am using Visual Studio .Net to make the transformation.
>
> > Best,
> > Michael
>
> --
>
> =A0 =A0 =A0 =A0 Martin Honnen --- MVP XML
> =A0 =A0 =A0 =A0http://JavaScript.FAQTs.com/

Thanks Martin,

I have to write code that will work on any nodenames/trees so the
above code will not work for me.

Best,
Michael