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[xsl] Re: The Solution (Was: Re: extracting sequences)

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A minor correction:

> When the above transformation is applied on your source.xml (corrected to
be
> well-formed!):

[wrong xml (one of the stylesheets was pasted here by mistake) snipped]

Below is the original (corrected) xml document:

<RDF xmlns:r="http://www.w3.org/TR/RDF/"
xmlns:d="http://purl.org/dc/elements/1.0/">
  <Node r:id="Root">
    <num> 1</num>
    <d:Title> Root</d:Title>
    <link r:resource="Root/a"></link>
    <link r:resource="Root/b"></link>
    <link r:resource="Root/c"></link>
  </Node>
  <Node r:id="Root/a">
    <num> 2</num>
    <d:Title> a</d:Title>
    <link r:resource="Root/a/d"></link>
    <link r:resource="Root/a/e"></link>
  </Node>
  <Node r:id="Root/b">
    <num> 3</num>
    <d:Title> b</d:Title>
    <link r:resource="Root/b/g"></link>
    <crosslink r:resource="f:Root/a/e"/>
  </Node>
  <Node r:id="Root/c">
    <num> 4</num>
    <d:Title> c</d:Title>
    <crosslink r:resource="h:Root/b/g"/>
  </Node>
  <Node r:id="Root/a/d">
    <num> 5</num>
    <d:Title> d</d:Title>
  </Node>
  <Node r:id="Root/a/e">
    <num> 6</num>
    <d:Title> e</d:Title>
  </Node>
  <Node r:id="Root/b/g">
    <num> 7</num>
    <d:Title> g</d:Title>
  </Node>
</RDF>

>
> the wanted result is produced:
>
> a d
> a e
> b f
> b g
> c h
>
>
> Hope this helped.
>
>
> Cheers,
>
> Dimitre Novatchev
> FXSL developer,
>
> http://fxsl.sourceforge.net/ -- the home of FXSL
> Resume: http://fxsl.sf.net/DNovatchev/Resume/Res.html
>
>
>
>
>  XSL-List info and archive:  http://www.mulberrytech.com/xsl/xsl-list
>
>




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