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

[Chad Chelius <cchelius@xxxxxxxxxxxxxxx>]

David,
Below is the XSLT that I am currently trying to use. The portion  
above the first comment creates the structure that I want except that  
it doesn't include the <Root> element that the <Story> element is  
contained in. Then when I add the instructions below the first  
comment to try to rename the elements, I lose the structure and tags  
that I want. My XML file structure is as such:

<Root>
	<Story>
		<articleTitle>Content</articleTitle>
		<articleSubTitle>Content</articleSubTitle>
		<author><authorFname>content</authorFname></author>

This is my stylesheet:

<xsl:template match="Story">
       <xsl:copy>
       <BB><TG>
<xsl:apply-templates select="articleTitle|articleSubTitle"/>
       </TG></BB>
       <xsl:apply-templates select="*[not(self::articleTitle or  
self::articleSubTitle)]"/>
       </xsl:copy>
  </xsl:template>
<!--Above this line creates the directory structure that I need-->
    <xsl:template match="Root">
        <DG>
            <xsl:apply-templates select="@*|node()"/>
        </DG>
    </xsl:template>
    <xsl:template match="Story">
        <D>
            <xsl:apply-templates select="@*|node()"/>
        </D>
    </xsl:template>

<chad/>

Chad Chelius
AGI Training
cchelius@xxxxxxxxxxxxxxx


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