neo4j cypher：查找带有没有特定传出关系的末端节点的所有路径吗？

Question

I want to use cypher to get to the following result: "Find all paths starting from node A only following relationships on which the property "percentage" ¹ is greater 50 and the end of path is a node with property 'type' = 1 and the end of path has no further relationships as specified before (percentage>50 ...)"

( ¹ : I'm considering to create a separate relationship-type "MY_RELATIONSHIP_50" for performance reasons)

My Cypher works fine so far:
start A = node(...)
match path = (A)-[rels:MY_RELATIONSHIP*]->(B)
where all(rel in rels where rel.percentage! > 50) and B.type = 1
return path

But I can't find a way to express " the end of path has no further relationships as specified before (percentage>50 ...) "

I tried to extend the where clause with " and not B-->C " but I did neither find out how to qualify with percentage > 50 .

Is there a way to do this?

Thank's a lot in advance =)

Answer 1

You are kind of looking for the longest path?

So either just sort descending by length and take the top n.

Otherwise something like:

start A = node(...)
match path = (A)-[rels:MY_RELATIONSHIP*]->(B)
where all(rel in rels where rel.percentage! > 50) and B.type = 1
AND NONE(r in extract(p in (B-[:MY_RELATIONSHIP]->()) : head(rels(p)) WHERE r.percentage > 50)
return path