Re: [xsl] Node selection based on parent attribute

[Joerg Heinicke <joerg dot heinicke at gmx dot de>]

must be <xsl:if test="..."/>

Regards,

Joerg

katharine wykes wrote:
> Still trying to get a better understanding of xslt.  Using the code you 
> specified below brings up a parser error:
> Attribute 'select' is invalid on 'xsl:if'.
>
> Any clues as to the cause of this?
> Cheers,
>
>
> > From: Wendell Piez <wapiez@mulberrytech.com>
> > Reply-To: xsl-list@lists.mulberrytech.com
> > To: xsl-list@lists.mulberrytech.com
> > Subject: Re: [xsl] Node selection based on parent attribute
> > Date: Fri, 30 Aug 2002 12:24:31 -0400
> >
> > Katharine,
> >
> > Another approach (besides the ones using for-each) is just to use the 
> > built-in descent of the tree, as in something like:
> >
> > <xsl:template match="menu">
> >   <xsl:if select="ancestor::menu[@id=$id']">
> >    <xsl:value-of select="@id"/>
> >    <!-- copies out this menu's @id if it has an ancestor
> >         menu with @id = $id -->
> >   </xsl:if>
> >   <xsl:apply-templates/>
> >   <!-- continues the tree traversal in case there are any below -->
> > </xsl:template>
> >
> > Make sure the parameter is set to the *value* of the id ('1', '6', 
> > whatever) whose descendants you want.
> >
> > Cheers,
> > Wendell

--

System Development
VIRBUS AG
Fon  +49(0)341-979-7419
Fax  +49(0)341-979-7409
joerg.heinicke@virbus.de
www.virbus.de


XSL-List info and archive:  http://www.mulberrytech.com/xsl/xsl-list