Re: [xsl] Omitting elements from output

["Joerg Heinicke" <joerg.heinicke@xxxxxx>]

Hello Rick,

in XSLT there is a built-in template. This gives you the text-value of all
nodes, for which you don't have a specific template. You have two
possibilities: Either changing <xsl:apply-templates/> to
<xsl:apply-templates select="Element1 | Element2"/>. Or you create another
for all other nodes, which does not give you the text-value: <xsl:template
match="*"/>.

Regards,

Joerg

> I have the follow structure in my stylesheet.
>
> <xsl:template match="/">
>     <xsl:apply-templates/>
> <xsl:template>
>
> <xsl:template match="Element1">
>     ...
> </xsl:template>
>
> <xsl:template match="Element2">
>     ...
> </xsl:template>
>
> When I transform the XML file, I only want the content of Element1 and
> Element2, but the content of the other elements is output as well. How do
I
> suppress the output of the other elements. Thanks.
>
> Rick Quatro


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