Question

在此cypher查询中，将返回与STATUS =“on”属性相互关联的节点之间的最长路径/路径，但我还想获得路径/路径的最后一个节点。

查询：

START n=node(*)
MATCH p=n-[rels:INCLUDE*]->m 
WHERE ALL (rel IN rels 
  WHERE rel.status='on') 
WITH COLLECT(p) AS paths, MAX(length(p)) AS maxLength 
RETURN FILTER(path IN paths 
  WHERE length(path)= maxLength) AS longestPaths

我该如何将其添加到查询中？感谢。

Answer 1

这会产生两个数组。第一个数组是每个路径中的最后一个项目，第二个是每个路径：

START n=node(*)
MATCH p=n-[rels:INCLUDE*]->m 
WHERE ALL (rel IN rels 
  WHERE rel.status='on') 
WITH COLLECT(p) AS paths, MAX(length(p)) AS maxLength 
WITH FILTER(path IN paths WHERE length(path)= maxLength) AS longestPaths
RETURN EXTRACT(path IN longestPaths | LAST(path)) as last, longestPaths

Answer 2

由于路径是集合，您可以应用LAST函数。

Answer 3

此示例用于从节点的每个分支中获取最后一个节点
与next_action关系

联系

MATCH p=(a:acct)-[:next_action*]->(c)
WITH COLLECT({node:c, l:length(p)}) AS paths, MAX(length(p)) AS maxLength, a.uid as uid
WITH [x IN paths WHERE x.l= maxLength] AS last_node
RETURN last_node

有些奇怪，在Neo4j GUI界面中运行此查询将带走所有最后的节点。从py2neu运行相同的查询不会。从py2neo运行时，以下修改将起作用

MATCH p=(a:acct)-[:next_action*]->(c)
WITH COLLECT({node:c, l:length(p)}) AS paths, MAX(length(p)) AS maxLength, a.uid as uid
WITH [x IN paths WHERE x.l= maxLength] AS last_node
WITH COLLECT(last_node) as last_nodes
RETURN last_nodes

Answer 4

从 Neo4J 4.X 开始，不再支持 FILTER()，因此可以将其替换为 REDUCE()。方法如下::

START n=node(*)
MATCH p=n-[rels:INCLUDE*]->m 
WHERE ALL (rel IN rels 
  WHERE rel.status='on') 
WITH COLLECT(p) AS paths, MAX(length(p)) AS maxLength 
// WITH FILTER(path IN paths WHERE length(path)= maxLength) AS longestPaths
WITH REDUCE( newPaths = [], path in paths |
            CASE
            WHEN length(path)= maxLength THEN newPaths + [path]
            END ) AS longestPaths
RETURN EXTRACT(path IN longestPaths | LAST(path)) as last, longestPaths

如何在neo4j中获取路径中的最后一个节点？

4 个答案: