[GraphHopper] Extract a single edge?

[Andreas Fischer]

1000x Thanks!!!

This solved the problem as you said:

EdgeExplorer ee = graph.createEdgeExplorer(edgeFilter);
ee.setBaseNode(nodeID);

while(ee.next()){
  resultSet.add(graph.getEdgeProps(ee.getEdge(), ee.getAdjNode()));
}


The only thing is that i still have no street names on all roads. I would
have checked the OSM ways for a streetname, but in my screenshot you can
see, that some edges are purple which means that it has a name and doesn't
has a name at the same time!

> what are you trying to solve? I'm not sure I can follow you as this all
is already implemented in GraphHopper ...
I simply try to get all edges that are in my direct environment.

> Using the latest version would give you GPS exact results and we make
sure no other (already solved) problems are involved.
It is not easy to update to the latest version, since i changed essential
things to achieve my goals.

Regards,
Andreas


2013/12/16 Peter K <peathal at yahoo.de>

>
> > Maybe i understand the problem... Could it be that an edge explorer
> > only iterates over edges of which the base node is the base node?
> > Makes sense somehow.
> > Now is the problem: How can i find an edge to which a given node is
> > the adjacent node?
>
> Not sure if I understand completely. Maybe the problem you are having is
> that with CH a node is only connected to nodes with a higher level but
> still you should get all edges. And since you don't want shortcut edges
> this cannot be the problem. Hmmh.
>
> Or maybe the API is unclear? 'base' and 'adjacent' node is dependent of
> how you traverse the graph, not e.g. if this direction is accessible for
> car. E.g. if you have a simple graph '1-2-3' and you do
>
> iter = explorer.setBaseNode(1)
> while(iter.next()) {
>     edgeIds.add(iter.getEdge());
> }
> => you'll only get one edge id with adjacent node '2'
>
> iter = explorer.setBaseNode(2)
> while(iter.next()) {
>     edgeIds.add(iter.getEdge());
> }
> => you'll get two edge ids with adjacent node '1' and '3' where the
> order is not defined
>
> Regards,
> Peter.
>
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