使用Neo4j Cypher获取没有第一个或最后一个节点的最长路径

Question

I have a forest of nodes and relationships. Similar to the following:

N1-sends->N2-sends->N3-sends->N4

N5-sends->N6-sends->N7-sends->N8

N9-sends->N10-sends->N11-sends->N12-sends->N13

I want to write a Cypher query that returns the 3 paths, without the first or last item. The nodes that are NOT at the beginning or end of the paths already have a property("middle", "true"), so that makes it easier.

A problem that I have encountered is that Cypher returns the path and every SUBSET of the path as well. For example it returns n10->n11-> and n11->n2, and n10->n11->n12, .... which is not what I want.

Instead I just want the results to be an array of 3, where inside each I have:

n2->n3

n6->n7

n10->n11->n12

and thats it.

The queries that I have came up with are: (first one has syntax errors):

START n=node(*) MATCH p=()-[*]->i-[*]->() WHERE has(i.middle) 
WITH COLLECT(p) AS pa, MAX(length(p)) AS maxLength, NODES(p) AS pn
FILTER(path IN pa WHERE length(path)=maxLength) AS longestPaths 
RETURN DISTINCT FILTER(x IN longestPaths WHERE exists(x.middle)) 

and

START n=node(*) MATCH p=()-[*]->i-[*]->() 
WHERE has(i.middle)  
RETURN DISTINCT filter(x IN NODES(p) WHERE exists(x.middle)) as O

The second one returns the paths without the first and last node, but it returns duplicated nodes, because it returns subsets of path as well..

Thank you.

Answer 1

This might do what you want:

MATCH (n)-[:sends*]->(m)
WHERE NOT ( ()-[:sends]->(n) OR (m)-[:sends]->() )
RETURN NODES(p)[1..-1] AS middleNodes;

The WHERE clause eliminates sub-paths (ie, paths that are part of a longer path). The NODES(p)[1..-1] syntax returns the second through next-to-last nodes in each path.