Re: [xsl] how to rearrange nodes based on a dependency graph?

[Mark Nahabedian <naha@xxxxxxxxxx>]

Gunther Schadow writes:
 > I still can't see what this would do. This isn't about simply
 > filtering out nodes with dependencies, it is filtering out
 > those nodes that had already been processed because of the
 > dependency graph walk. The only way I can see to do this is
 > to somewhere keep a list of the processed nodes, but how
 > can I do that with XSL? (I guess I can't). The only alternative
 > would be to decide for every node if it would have been in
 > the dependency graph of any prior node. But that would mean
 > that all dependency graphs need to be always chased for every
 > node, that's kind of mind-boggling.

As I understand it, you're using the dependency information to
generate the required contents in the order dictated by the
dependecies, but that your getting duplicates of those items that are
depended on more than once.  The items have some unique id by which
they are referred to in the dependency relationships.

Suppose you use the dependency information to first generate a list of
the items (r their ids) to be included in the output, and you then
process the resulting list to only include an item if there's not an
identical preceeding sibling?

I think that for this approach you'd either need to do two sequential
transformations or use the node-set extension function.




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