Hey,
> Request is working like a charm and is fast (around 6 objects are checked
> in my 2M node 6M rel DB). I have to check the result with the corresponding
> RDBMS KPI.
Nice.
> Thanks again !
No problem.
> I definitively have to take more time to learn Gremlin !
You are more than welc
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> @Marko I am using neo4j 1.4 server so I guess I am running gremlin 1.1.
> It would be cool if you can send me the compatible query when you have time.
Good, cause time is all I have.
m = [:]
g.idx('id')[[object_type:'A']].transform{it.out('USES'){it.object_type.equals('B')}.c
@Marko I am using neo4j 1.4 server so I guess I am running gremlin 1.1.
It would be cool if you can send me the compatible query when you have time.
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Hey,
> @Marko thanks. I just tried to do it with Gremlin but I have not your
> knowledge of gremlin. I will try to do that.
If you use Neo4j Server 1.4, then you will be in Gremlin 1.1 land. I don't know
what you are rolling with (embedded, server, etc.), but if you want the query
in Gremlin 1.
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Hey guys,
>> I have 2 kind of nodes (let's say A and B) which are related using a
>> relation of type 'USES'.
>> I want to count all the type A nodes which have only 1 relation to B type
>> node and all the type A nodes which have more than one relation to B type
>> node.
>>
>> I am using neo4
Guillaume,
sorry, your mail seem to have slipped through the cracks.
Aggregate functions in the where clause don't work.
We intend to add a having clause to cypher in the 1.5 timeframe.
Then it should be possible to do something like this.
> start A=(id,object_type,"A") match (A)-[r:USES]->(B)
Neo4jists,
Was my question dumb or was it really not clear ?
Cheers
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Hi,
I have 2 kind of nodes (let's say A and B) which are related using a relation
of type 'USES'.
I want to count all the type A nodes which have only 1 relation to B type node
and all the type A nodes which have more than one relation to B type node.
I am using neo4j server and I am currently
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