Indeed, that's exactly what I did to come up with next and previous siblings (I wasn't using same-name siblings for other reasons). It would have been nice to have some sort of shortcut to avoid iterating over all siblings.
I notice that NodeIterator has a getPosition() method - if that position were available to the node itself, one could skip() directly to the previous and next siblings via the NodeIterator from node.getParent().getNodes(). -Brian On 7/26/07, Christoph Kiehl <[EMAIL PROTECTED]> wrote: > > Brian Thompson wrote: > > > In my case, it was a web app where each sibling node corresponded to a > page > > being displayed, and we needed a paging mechanism (the pages formed a > > training course). > > Since a child node does not know anything about its siblings you will have > to > ask the parent which means essentially you have to call > childNode.getParent().getNodes() and use the resulting iterator to find > your > specific child node and then identify its neighbors. > If you use same name siblings you could also use something like this: > > <snip>
