Hey I thought I'd try this out. Got the new kernel and downloaded the new index component snapshot but can't figure out how to use it. Perhaps I'm just dense. Looks like the only class in the Javadoc is an abstract class so I don't know how to instantiate a new relationship index. Have an example?
Thanks, Jeff Klann On Fri, Jul 30, 2010 at 10:19 AM, Mattias Persson <[email protected] > wrote: > The latest snapshots of things can be found at http://m2.neo4j.org/ > and this component (jar-file) can be found in > http://m2.neo4j.org/org/neo4j/neo4j-lucene-index/0.1-SNAPSHOT/ > > 2010/7/30, Arijit Mukherjee <[email protected]>: > > Thanx Mattias. Can I download a tar.gz or zip file from somewhere? I'm > > not using Maven in my projects yet...I mean I'm not very comfortable > > with it. > > > > Arijit > > > > On 30 July 2010 17:33, Mattias Persson <[email protected]> > wrote: > >> Looping through relatiomships manually is the way to go. However > >> there's a new component in > >> https://svn.neo4j.org/laboratory/components/lucene-index/ which can > >> index relationships and do fast lookups on whether or not a > >> relationship (with a certain attribute even) exists between two nodes. > >> > >> You'll need to go with the latest kernel then as well (as seen in > >> https://svn.neo4j.org/laboratory/components/lucene-index/pom.xml). > >> > >> 2010/7/30, Arijit Mukherjee <[email protected]>: > >>> Hi All > >>> > >>> I have a requirement where I must check if there is an already > >>> existing relationship between two nodes (say N1 and N2). Right now, > >>> I'm doing it as follows: > >>> > >>> boolean found = false; > >>> final Iterable<Relationship> currentRels = > >>> N1.getRelationships(RelTypes.KNOWS, Direction.OUTGOING); > >>> for (Relationship rel : currentRels) { > >>> found = rel.getEndNode().equals(N2); > >>> if (found) { > >>> do something - like add some property to the existing > >>> relationship; > >>> break; > >>> } > >>> } > >>> if (!found) { > >>> create new relationship between N1 and N2; > >>> } > >>> > >>> This means, for a high volume of data, all the relations going out of > >>> N1 will be retrieved and checked - and this seems costly. I'm using > >>> the 1.0 API, and wasn't able to find anything that would directly > >>> check whether N1 has an outgoing relationship with N2 - like > >>> N1.hasRelationship(N2, Direction.OUTGOING) - or something similar. I > >>> think there was a similar mail sometime ago. Has there been any > >>> updates lately which allows such checks? Or, is there any other direct > >>> way to do this with the 1.0 API? > >>> > >>> Regards > >>> Arijit > >>> > >>> -- > >>> "And when the night is cloudy, > >>> There is still a light that shines on me, > >>> Shine on until tomorrow, let it be." > >>> _______________________________________________ > >>> Neo4j mailing list > >>> [email protected] > >>> https://lists.neo4j.org/mailman/listinfo/user > >>> > >> > >> > >> -- > >> Mattias Persson, [[email protected]] > >> Hacker, Neo Technology > >> www.neotechnology.com > >> _______________________________________________ > >> Neo4j mailing list > >> [email protected] > >> https://lists.neo4j.org/mailman/listinfo/user > >> > > > > > > > > -- > > "And when the night is cloudy, > > There is still a light that shines on me, > > Shine on until tomorrow, let it be." > > _______________________________________________ > > Neo4j mailing list > > [email protected] > > https://lists.neo4j.org/mailman/listinfo/user > > > > > -- > Mattias Persson, [[email protected]] > Hacker, Neo Technology > www.neotechnology.com > _______________________________________________ > Neo4j mailing list > [email protected] > https://lists.neo4j.org/mailman/listinfo/user > _______________________________________________ Neo4j mailing list [email protected] https://lists.neo4j.org/mailman/listinfo/user
