如何在Neo4j中获取路径中的最后一个节点？

Question

在这个密码查询中，将返回具有STATUS="on"属性关系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

我该如何将它添加到查询中？谢谢。

- fereshteh

Answer 1

这将生成两个数组。第一个数组是每条路径中的最后一项，第二个是每条路径：

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

- LameCoder

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@LameCoder 有没有办法获取最后一个节点，而不是路径？我得到的是一些 JSON 集合，而不是节点？ - Aman Gupta

要将此内容翻译为Neo4j 2.x，请将LAST(path)更改为LAST(nodes(path))，如下所示的2.x注释。无论如何，这对我有效。 - Andrew Lank

Answer 2

既然路径是一个集合，你可以应用LAST函数。

- Stefan Armbruster

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从neo4j 2.0开始，路径不再是一个集合，因此您不能直接在其上运行LAST。 - Tristan Reid

你可以使用 last(nodes(mypath))。它的意思是获取路径 mypath 中的最后一个节点。 - Stefan Armbruster

Answer 3

这个示例是用于获取与next_action关系相连的每个节点分支的最后一个节点。

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

在Neo4j GUI界面中运行此查询会显示所有最后的节点，但从py2neo运行相同的查询则不会。在py2neo中运行时，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

- user3827333

Answer 4

从Neo4J 4.X开始，FILTER()不再受支持，因此可以用REDUCE()替换。 以下是替换的方式：

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

- Tanmay Banerjee.

Answer 5

对于任何想要获取路径中最后一个节点的人：

如果最后一个节点具有相同的类型：

MATCH (c:Tweet{ id: 1 })-[:REPLY_TO*]->(target)
RETURN last(collect(target))

如果最后一个节点类型不同：

MATCH (c:Comment{ id: 1 })-[:COMMENT_TO*]->(post:HeychainPost)
RETURN post

- Aarni Alasaarela