如何获取neo4j路径中的最后一个节点？答案

【问题标题】：how to get the last node in path in neo4j?如何获取neo4j路径中的最后一个节点？
【发布时间】：2013-11-15 08:26:55
【问题描述】：

在此密码查询中，将返回与 STATUS="on" 属性相关的节点之间的最长路径/路径，但我还想获取路径/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

我应该如何将它添加到查询中？谢谢。

【问题讨论】：

标签： neo4j cypher path-finding

【解决方案1】：

这将给出两个数组。第一个数组是每个路径中的最后一项，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

【讨论】：

@LameCoder 有没有办法获取最后一个节点而不是路径我得到一些 JSON 集合而不是节点？
要将其转换为 Neo4j 2.x，请将下面的 2.x cmets 中的 LAST(path) 更改为 LAST(nodes(path))。无论如何，这对我有用。

【解决方案2】：

由于路径是一个集合，您可以应用LAST 函数。

【讨论】：

从 neo4j 2.0 开始，路径不是集合，不能直接在其上运行 LAST。
你可以使用last(nodes(mypath))

【解决方案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

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

有点奇怪，在 Neo4j GUI 界面中运行这个查询会带来所有最后的节点。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

【讨论】：

【解决方案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

【讨论】：