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使用Cypher查询Neo4j中的路径相交_Neo4j_Path_Cypher_Intersection - Fatal编程技术网

使用Cypher查询Neo4j中的路径相交

使用Cypher查询Neo4j中的路径相交,neo4j,path,cypher,intersection,Neo4j,Path,Cypher,Intersection,使用Cypher(Neo4j)执行此查询: 它返回属于特定组的所有可能路径(组只是对节点进行分类的属性),我正在努力获得一个查询,该查询返回两个路径集合之间的公共路径。应该是这样的: MATCH p=(g:Node)-[:FOLLOWED_BY *2..2]->(g2:Node) WHERE g.group=10 AND g2.group=10 MATCH p=(g3:Node)-[:FOLLOWED_BY *2..2]->(g4:Node) WHERE g3.group=15

使用Cypher(Neo4j)执行此查询:

它返回属于特定组的所有可能路径(组只是对节点进行分类的属性),我正在努力获得一个查询,该查询返回两个路径集合之间的公共路径。应该是这样的:

MATCH p=(g:Node)-[:FOLLOWED_BY *2..2]->(g2:Node) 
WHERE g.group=10 AND g2.group=10
MATCH p=(g3:Node)-[:FOLLOWED_BY *2..2]->(g4:Node) 
WHERE g3.group=15 AND g4.group=15 
RETURN INTERSECTION(path1, path2)

当然是我编的。目标是获得两个查询之间的所有公共路径。

您的2个
匹配的开始/结束节点具有不同的
s,因此它们永远找不到公共路径

因此,当您询问“公共路径”时,我假设您实际上希望找到共享的中间节点(在两组3节点路径之间)。如果是这样,此查询应该可以工作:

MATCH p1=(g:Node)-[:FOLLOWED_BY *2]->(g2:Node) 
WHERE g.group=10 AND g2.group=10
WITH COLLECT(DISTINCT NODES(p1)[1]) AS middle1
MATCH p2=(g3:Node)-[:FOLLOWED_BY *2]->(g4:Node) 
WHERE g3.group=15 AND g4.group=15 AND NODES(p2)[1] IN middle1
RETURN DISTINCT NODES(p2)[1] AS common_middle_node;
MATCH p1=(g:Node)-[:FOLLOWED_BY *2]->(g2:Node) 
WHERE g.group=10 AND g2.group=10
WITH COLLECT(DISTINCT NODES(p1)[1]) AS middle1
MATCH p2=(g3:Node)-[:FOLLOWED_BY *2]->(g4:Node) 
WHERE g3.group=15 AND g4.group=15 AND NODES(p2)[1] IN middle1
RETURN DISTINCT NODES(p2)[1] AS common_middle_node;