CODEWITHSUNDEEP
rubyd3.jsneo4jneo4j.rbvisualize
f7n
f7n

Reputation: 1684

How to visualise the relationship between two nodes in Neo4j using d3.js?

I've wrote a cypher query to show the relationship between two movie nodes:

MATCH (m1:Movie)-[r*1..2]-(m2:Movie)
where m1.movieID = '1' AND m2.movieID = '2'
RETURN r
LIMIT 25

It's returns this in the Neo4j browser, and this is exactly what I want to create in d3.js. Eventually the movieIDs will be parameters, so the results will be different for each pair of movies.

enter image description here

I've been following the Neo4j-core Ruby example seen here to create my own graph, but my problem is that I can't RETURN common nodes between the two moves, i.e Tag/Country nodes.

How should I solve this problem? Should I create a query to return the two movies, and another query to return the common nodes between them? How would I write this latter query? I'm really stuck on how I can visualise this relationship...

Upvotes: 1

Views: 807

Answers (2)

Brian Underwood
Brian Underwood

Reputation: 10856

You could return the movies with the relationships:

MATCH (m1:Movie)-[r*1..2]-(m2:Movie)
WHERE m1.movieID = '1' AND m2.movieID = '2'
RETURN m1, m2, r
LIMIT 25

Though that might give you duplication. If you're going to be displaying this with D3 probably what you want is all of the relationships involved and their start/end nodes. To do that you could do:

MATCH (m1:Movie)-[rels*1..2]-(m2:Movie)
WHERE m1.movieID = '1' AND m2.movieID = '2'
RETURN rels
UNWIND rels AS r
WITH DISTINCT r AS rel
WITH startnode(rel) AS startnode, endnode(rel) AS endnode, rel
LIMIT 25

Generally what you want to end up with is a collection of nodes and relationships involved. So with the result of that query you might do:

result = neo4j_session.query(query_string)

data = {nodes: [], relationships: []}
result.each do |row|
  data[:nodes] << row.startnode
  data[:nodes] << row.endnode
  data[:relationships] << row.rel
end

data[:nodes].uniq!

Upvotes: 0

FrobberOfBits
FrobberOfBits

Reputation: 18002

Perhaps you should try matching the path rather than the relationships as you are right now. Something like:

MATCH p=(m1:Movie { movieID: 1 } )-[r*1..2]-(m2:Movie { movieID: 2 } )
RETURN p;

You can then use collection functions on the path object p to do whatever you like with it. For example, if you want to get the inner nodes, you can use nodes(p) and simply remove the first and last nodes (which would be m1 and m2) and get the nodes along the path.

Upvotes: 2

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