Re: [xsl] Re: xsl-list Digest 1 Jun 2006 05:10:01 -0000 Issue 793

[Chad Chelius <cchelius@xxxxxxxxxxxxxxx>]

David,
I think this is making more sense now. The first part of the template  
below, creates the parent elements like we had discussed before, and  
the tags for the child elements of <Story> in this case don't start  
to show up until I tell them to like in the second part of the  
template. One thing that I have run into is the fact that I need to  
rename the <Story> element after or during processing. How do I  
accomplish something like that?

   <xsl:template match="Story">
        <xsl:copy>
            <BB><TG><TI>
                <xsl:apply-templates select="articleTitle| 
articleSubTitle"/>
            </TI></TG></BB>
            <xsl:apply-templates select="*[not(self::articleTitle or  
self::articleSubTitle)]"/>
        </xsl:copy>
    </xsl:template>

<-- Second part -->

    <xsl:template match="Root">
        <DG>
            <xsl:apply-templates select="@*|node()"/>
        </DG>
    </xsl:template>

chad


On Jun 1, 2006, at 12:05 PM, David Carlisle wrote:

> > I need all of the other elements to come over as
> > well and I need to rename those elements.
> yes that's what the
>  <xsl:apply-templates select="*[not(self::title or self::subtitle)]"/>
> is doing, applying your templates to all the child elements.
> Sounds like you don't want document to go to document though so change
>  <xsl:copy>  to <D>
>
> and when I said
>
> > you just need a standard identity template,
> read that as "you just need the templates doing whatever  
> transformation
> they are doing"
>
> David