Re: [xsl] using xsl:result-document in a non-list situation

["Andrew Welch" <andrew.j.welch@xxxxxxxxx>]

2008/8/19 Mark Wilson <mark@xxxxxxxxxxxx>:
> All of the examples for using <xsl:result-document> to produce multiple
> files (at least those that I've seen) write to a new file each time they
> encounter a new instance of a controlling element structure in a 'list' and
> use "position()" to help name the file. What technique do I apply to a
> document containing a <List> of <Items> such as the one shown below? Here
> the controlling element is not simply another <Year> element, but <Year>
> elements with differing content: 1998, 1999 and so on, with each year
> repeated within the source  document many times. Ideally, the output would
> be a series of files, one for each different <Year> content (1998.xml,
> 1999.xml) and each sorted internally by <IssueNumber> and <Page>.
>
> I have not been able to see how the stylesheet should be constructed. Any
> hints would be appreciated.
> Thanks,
> Mark
>
> Source example:
> <List>
>   <Item>
>       <Data>
>           ...
>       </Data>
>       <Article>
>           <Year>1999</Year>
>           <IssueNumber>1</IssueNumber>
>           <Page>6</Page>
>       </Article>
>   </Item>
>   <Item>
>       .......
>   </Item>
> </List>

How about:

<xsl:for-each-group select="Item" group-by="Article/Year">
  <xsl:result-document href="{current-grouping-key()}.ext">
    <xsl:for-each select="current-group()">
      <xsl:sort select="Article/IssueNumber" data-type="number"/>
      <xsl:sort select="Article/Page" data-type="number"/>
      ......


cheers
-- 
Andrew Welch
http://andrewjwelch.com
Kernow: http://kernowforsaxon.sf.net/