Well so lastRelationship() isnt the right one method I am looking for
and I have to look for highest id manually.
But I have problem that if I traverse in the way like Michael suggested:
Iterator<Node> i =
Traversal.description().relationships(RelationshipTypes.TICKET_QUEUE,
Direction.BOTH).relationships(RelationshipTypes.TICKET_STATUS,
Direction.BOTH).
relationships(RelationshipTypes.TICKET_TIMETAKEN,
Direction.BOTH).
traverse(ticketNode).nodes().iterator();
The iterator gives me only 6 ancestors. But in Neoclipse I can see much
more ancestor nodes which is right. Why this traverse gives me bad
result? I need all nodes which are directly connected thought
Relatioshiptypes TICKET_STATUS,TICKET_TIMETAKEN,TICKET_QUEUE with
ticketNode...
Dne 4.4.2011 13:59, Mattias Persson napsal(a):
> 2011/4/4 Matěj Plch<plchm...@fit.cvut.cz>
>
>> Is it possible to use
>>
>> *Path.lastRelationship*()
>>
>> ?
>> How does it take last Relationship? According to id, or how the graph is
>> traversed?
>>
> It returns the last relationships in the current path, i.e. where the
> traverser is a.t.m. So it already has a reference to it and just returns it.
>
>> Dne 26.3.2011 19:35, Michael Hunger napsal(a):
>>> Sure, if the tree from your root node is just a cluster that is not
>> connected anywhere else (with those 3 relationship-types) it should be as
>> simple as.
>>> (Just written from my head, so please check the correct syntax).
>>>
>>>
>> Traversal.description().relationship(T1,OUTGOING).relationship(T2,OUTGOING).relationship(T3,OUTGOING).traverse(rootNode);
>>> That returns an iterator of all paths going from your root node.
>>>
>>> You can limit the nodes with .uniqueness() and then add the path's
>> (path.nodes()) to a set to collect all nodes.
>>> For getting the one with the highest id, you can use
>> java.util.Collections.max(collection, new Comparator<Node>(){});
>>> How big is your tree?
>>>
>>> Something like that should be in Graph-Algo perhaps as "subgraph" or
>> "tree".
>>> HTH
>>>
>>> Michael
>>>
>>> Am 26.03.2011 um 19:26 schrieb Matěj Plch:
>>>
>>>> Thank you for so fast answer.
>>>> I will look at it. I have milestone tomorrow so dont have a lot of
>>>> time=) and have never worked with Groovy.
>>>> Well so there isnt any simple method how to do it in classic neo4j Java
>> API?
>>>> Dne 26.3.2011 19:16, Saikat Kanjilal napsal(a):
>>>>> You can do all of these things using gremlin and pipes. Check out
>> github for more details.
>>>>> Sent from my iPhone
>>>>>
>>>>> On Mar 26, 2011, at 11:13 AM, Matěj Plch<plchm...@fit.cvut.cz>
>> wrote:
>>>>>> Hi, I have some graph and, part of it is a tree. I simple get root of
>>>>>> this tree through id. How to simple tranverse only tree under this
>> root
>>>>>> node? From root goes three unique type relationship to three unique
>>>>>> group type nodes. Under this three nodes are a lot of nodes. And I
>> need
>>>>>> to write a method which gives me all nodes under that group node.
>>>>>> Second question is if its possible ho to get from this group noe with
>>>>>> the highest id (last added).
>>>>>> Matěj Plch
>>>>>>
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