I had there an error... My fault. Your code is working fantastic. Thank
you so much. I hope the last question: is it possible to add some
parameter to exclude the start node?
Dne 6.4.2011 09:13, Mattias Persson napsal(a):
> I'm not fully aware of your domain layout, but maybe add this:
>
> .uniqueness( Uniqueness.RELATIONSHIP_GLOBAL )
>
> to your traversal description. The default (NODE_GLOBAL) may end up "hiding"
> some of your nodes depending on your graph layout.
>
> 2011/4/5 Matěj Plch<plchm...@fit.cvut.cz>
>
>> 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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