On May 1, 4:11=A0am, Martin Honnen <mahotr...@yahoo.de> wrote:
> time2...@gmail.com wrote:
> > I want to transform the XML so that the result is the same except one
> > of the c nodes is always returned first in the response based on a
> > specfic id element (see below).
>
> Pass in the id of the 'c' element as a parameter and then use the
> identity transformation template plus a template for 'b' elements that
> ensures that the 'c' children are processed in the order you want:
>
> <xsl:stylesheet
> =A0 =A0xmlns:xsl=3D"http://www.w3.org/1999/XSL/Transform"
> =A0 =A0version=3D"1.0">
>
> =A0 =A0<xsl:param name=3D"id" select=3D"3"/>
>
> =A0 =A0<xsl:strip-space elements=3D"*"/>
>
> =A0 =A0<xsl:output method=3D"xml" indent=3D"yes"/>
>
> =A0 =A0<xsl:template match=3D"@* | node()">
> =A0 =A0 =A0<xsl:copy>
> =A0 =A0 =A0 =A0<xsl:apply-templates select=3D"@* | node()"/>
> =A0 =A0 =A0</xsl:copy>
> =A0 =A0</xsl:template>
>
> =A0 =A0<xsl:template match=3D"b">
> =A0 =A0 =A0<xsl:copy>
> =A0 =A0 =A0 =A0<xsl:apply-templates select=3D"@*"/>
> =A0 =A0 =A0 =A0<xsl:apply-templates select=3D"c[@id =3D $id]"/>
> =A0 =A0 =A0 =A0<xsl:apply-templates select=3D"c[not(@id =3D $id)]"/>
> =A0 =A0 =A0</xsl:copy>
> =A0 =A0</xsl:template>
>
> </xsl:stylesheet>
>
> --
>
> =A0 =A0 =A0 =A0 Martin Honnen --- MVP XML
> =A0 =A0 =A0 =A0http://JavaScript.FAQTs.com/

This worked perfectly, thanks!