Re: [Neo4j] How to get both nodes and relationship in one traverse?
You can traverse only relatioships and call getStartNode() getEndNode()... Dne 12.4.2011 03:37, Brendan napsal(a): Hi, I had setup one traversal description but it seems to me that each time of traverse can only get one of three outcome, nodes, relationships, or paths. How to get more than one result traverse? Now, I sent two requests to get both nodes and relationships. But there is a chance that between the requests there are some changes in the database. Regards, Brendan Sent from my iPad ___ Neo4j mailing list User@lists.neo4j.org https://lists.neo4j.org/mailman/listinfo/user ___ Neo4j mailing list User@lists.neo4j.org https://lists.neo4j.org/mailman/listinfo/user
Re: [Neo4j] simple traverse of tree
And how it is possible to traverse parallel edges? I spent an hour
trying how to do it... I have some node and this node can be connected
with other node with more incoming edges. And I'm not able to traverse
all this edges...
IteratorRelationship rels =
Traversal.description().breadthFirst().evaluator(Evaluators.excludeStartPosition()).relationships(RelationshipTypes.TICKET_WORKER,
Direction.INCOMING).traverse(workerNode).relationships().iterator();
This traverse me only unique connections (not parallel)...
Dne 6.4.2011 23:49, Mattias Persson napsal(a):
Just as a note (and as the javadoc says) you can have multiple
evaluators in a traversal description.
2011/4/6 Michael Hungermichael.hun...@neotechnology.com:
sure
...
.evaluator(Evalutators.excludeStartPosition())
...
Cheers
Michael
Am 06.04.2011 um 20:15 schrieb Matěj Plch:
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 Plchplchm...@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:
IteratorNodei =
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 Plchplchm...@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 ComparatorNode(){});
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 Plchplchm...@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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Re: [Neo4j] simple traverse of tree
Sorry for the post. .uniqueness(Uniqueness.NONE) made the trick.
Dne 9.4.2011 13:19, Matěj Plch napsal(a):
And how it is possible to traverse parallel edges? I spent an hour
trying how to do it... I have some node and this node can be connected
with other node with more incoming edges. And I'm not able to traverse
all this edges...
IteratorRelationship rels =
Traversal.description().breadthFirst().evaluator(Evaluators.excludeStartPosition()).relationships(RelationshipTypes.TICKET_WORKER,
Direction.INCOMING).traverse(workerNode).relationships().iterator();
This traverse me only unique connections (not parallel)...
Dne 6.4.2011 23:49, Mattias Persson napsal(a):
Just as a note (and as the javadoc says) you can have multiple
evaluators in a traversal description.
2011/4/6 Michael Hungermichael.hun...@neotechnology.com:
sure
...
.evaluator(Evalutators.excludeStartPosition())
...
Cheers
Michael
Am 06.04.2011 um 20:15 schrieb Matěj Plch:
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 Plchplchm...@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:
IteratorNode 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 Plchplchm...@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 ComparatorNode(){});
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 Plchplchm...@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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Re: [Neo4j] simple traverse of tree
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 Plchplchm...@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:
IteratorNode 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 Plchplchm...@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 ComparatorNode(){});
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 Plchplchm...@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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Re: [Neo4j] simple traverse of tree
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:
IteratorNode 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 Plchplchm...@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 ComparatorNode(){});
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 Plchplchm...@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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[Neo4j] Glassfish problem
Hi! So we deployed our application, but we get Caused by: java.lang.IllegalStateException: Unable to lock store ... SEVERE: 0-database/neostore], this is usually caused by another Neo4j kernel already running in this JVM for this particular store Is there any best practise how to avoid this? The database is opened or created from class with is Singleton. Can this be the problem? I googled this problem, but no solution found. Matěj Plch ___ Neo4j mailing list User@lists.neo4j.org https://lists.neo4j.org/mailman/listinfo/user
[Neo4j] simple traverse of tree
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 ___ Neo4j mailing list User@lists.neo4j.org https://lists.neo4j.org/mailman/listinfo/user
Re: [Neo4j] simple traverse of tree
Well maybe the best one to show the graph:
https://webdev.fit.cvut.cz/~plchmate/mi-w20/neoclipse.png
https://webdev.fit.cvut.cz/%7Eplchmate/mi-w20/neoclipse.png
There is a ticket node:1001 and I would like to get all nodes under for
example node Status_Events. Now there is only one node, in real there
will be more nodes. But not so much (not more than 100 Im quite sure).
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 ComparatorNode(){});
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 Plchplchm...@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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[Neo4j] Dynamic GEXF node attributes in neo4j database
Hi! Do you have any experience how to store changing attributes of node? It's like to build a table of changes for each changed node attribute. In SQL datastore quite simple, but in no-SQL? And one more question. Is it possible with some easy call though Java API how to get number of nodes and edges stored in the database? Or I have to traversal all graph? Thank you Matej Plch ___ Neo4j mailing list User@lists.neo4j.org https://lists.neo4j.org/mailman/listinfo/user
Re: [Neo4j] neoclipse ubuntu 10.04 problem
Sure,
ok when Im working with db created with Java API, just call:
private static void getNode() {
int ticketId1 = 1;
GraphDatabaseService graphDb = new
EmbeddedGraphDatabase(var1/graphdb);
Transaction tx = graphDb.beginTx();
try {
Node node = graphDb.getNodeById(ticketId1);
int id = Integer.parseInt(node.getProperty(id).toString());
System.out.println(node1 id: + id);
} finally {
tx.finish();
graphDb.shutdown();
}
}
Get right result...
Wanted to open it with neoclipse, get that error. Wanted to open it with
Gephi with neo4j support get error with version of neo4j library problem
(tried with 1.03.M4 and the same error)...
But it I get error with also when I'm creating absilutelz new database
from Neoclipse... There can be only one neo4j db instance in system? How
can I see, if its open or not? Netstat, ps aux?
Thank you for your support...
Dne 19.3.2011 09:15, Peter Neubauer napsal(a):
Matej,
Are you sure your database is shut down properly from your java code
before opening it with Neoclipse ?
Try to just open and close it and make sure there are no exceptions.
HTH
/peter
On Saturday, March 19, 2011, Matěj Plchplchm...@fit.cvut.cz wrote:
Hi,
I'm running Ubuntu 10.04 (2.6.32-30) 32bit and have problem with
neoclipse. I downloaded neoclipse. I wanted to open some db made from
Java application, but I got some errors. So I set path to new db and
clicked to connect/start database. And got:
./neoclipse
19.3.2011 2:02:27 org.neo4j.neoclipse.graphdb.GraphDbServiceManager logInfo
INFO: Starting GraphDbServiceManager
19.3.2011 2:02:41 org.neo4j.neoclipse.graphdb.GraphDbServiceManager logInfo
INFO: trying to start/connect ...
19.3.2011 2:02:41 org.neo4j.neoclipse.graphdb.GraphDbServiceManager logInfo
INFO: connected to embedded neo4j
java.util.concurrent.ExecutionException:
java.lang.IllegalStateException: Can't start new database: the old one
isn't shutdown properly.
In the info window:
java.lang.IllegalStateException: Can't start new database: the old one
isn't shutdown properly.: Can't start new database: the old one isn't
shutdown properly.
There are no errors in messages.log. The db was created.
How can I get rid of this problem?
Running neo4j and also neoclipse version 1.2.
$ java -version
java version 1.6.0_24
Java(TM) SE Runtime Environment (build 1.6.0_24-b07)
Java HotSpot(TM) Server VM (build 19.1-b02, mixed mode)
Best regards
Matěj Plch.
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[Neo4j] neoclipse ubuntu 10.04 problem
Hi, I'm running Ubuntu 10.04 (2.6.32-30) 32bit and have problem with neoclipse. I downloaded neoclipse. I wanted to open some db made from Java application, but I got some errors. So I set path to new db and clicked to connect/start database. And got: ./neoclipse 19.3.2011 2:02:27 org.neo4j.neoclipse.graphdb.GraphDbServiceManager logInfo INFO: Starting GraphDbServiceManager 19.3.2011 2:02:41 org.neo4j.neoclipse.graphdb.GraphDbServiceManager logInfo INFO: trying to start/connect ... 19.3.2011 2:02:41 org.neo4j.neoclipse.graphdb.GraphDbServiceManager logInfo INFO: connected to embedded neo4j java.util.concurrent.ExecutionException: java.lang.IllegalStateException: Can't start new database: the old one isn't shutdown properly. In the info window: java.lang.IllegalStateException: Can't start new database: the old one isn't shutdown properly.: Can't start new database: the old one isn't shutdown properly. There are no errors in messages.log. The db was created. How can I get rid of this problem? Running neo4j and also neoclipse version 1.2. $ java -version java version 1.6.0_24 Java(TM) SE Runtime Environment (build 1.6.0_24-b07) Java HotSpot(TM) Server VM (build 19.1-b02, mixed mode) Best regards Matěj Plch. ___ Neo4j mailing list User@lists.neo4j.org https://lists.neo4j.org/mailman/listinfo/user
[Neo4j] java application server or neo4j server?
Hi! I'm new in neo4j (and also NonSQL databases). We are developing application which will fetch some information periodically, save it in nodes and edges in neo4j database and then import it to GEXF format to use our data in Gephi. My question is if it's possible to write this application like plugin (write another services) for Neo4j server or better approach is to run it on some application server. I'm probably little bit confused about for what is Neo4j server determined and what all it's possible to do. Thank you for your answers Matej Plch. ___ Neo4j mailing list User@lists.neo4j.org https://lists.neo4j.org/mailman/listinfo/user
Re: [Neo4j] java application server or neo4j server?
Thank you for your answer. So Neo4j server serves mainly for remote access to neo4j database but you don't recommend me to use it like application server, right? And plugins are just for extension funcionality of neo4j server's REST API. Dne 10.3.2011 13:38, Jim Webber napsal(a): Hi Matěj, Neo4j has traditionally been an embedded database, so if you just want to host it in an existing application (e.g. in a Web app you already run on an app server) then that's probably a good choice. If you need to host the database elsewhere on the network (or you're not running on the JVM) then Neo4j server supports a REST interface through which you can interact with the database. From your email, it sounds like the embedded database might be best for you. Jim ___ Neo4j mailing list User@lists.neo4j.org https://lists.neo4j.org/mailman/listinfo/user ___ Neo4j mailing list User@lists.neo4j.org https://lists.neo4j.org/mailman/listinfo/user
