在路径上的所有节点上执行MATCH [英] Performing a MATCH on all nodes on a path

问题描述

是否可以在MATCH模式> ALL函数(使用v1.8)?

Is it somehow possible to use a MATCH pattern inside the ALL function (using v1.8)?

我要执行的操作如下:我正在MATCH设置路径p = (a)-->(b)-->(c)-->(d).但是，沿该路径的所有节点都必须具有来自某个节点的附加传入关系r.让我试着用ASCII明确这一点:

What I am trying to do is the following: I am MATCHing a path p = (a)-->(b)-->(c)-->(d). However, all nodes along this path must have an additional incoming relationship r from some node. Let me try to make this clear in ASCII:

(a)-->(b)-->(c)-->(d)
       ^     ^     ^
       |r    |r    |r
      ( )   ( )   ( )

我可以以某种方式使用ALL函数还是必须添加其他如下的MATCH模式:

Can I somehow use the ALL function for that or do I have to add additional MATCH patterns like this:

START ...
MATCH (a)-->(b)-->(c)-->(d)..., ()-[:r]->(b), ()-[:r]->(c), ...
RETURN ...

更新:

Update:

以下是 Neo4j控制台中的示例:

START n=node(0) 
CREATE (a), (b), (c), (d), (e),
n-[:rel1]->a, n-[:rel1]->b, n-[:rel1]->d, n-[:rel1]->e,
a-[:rel2]->b-[:rel3]->d, a-[:rel2]->c-[:rel3]->e

START n=node(0) 
MATCH n -[:rel1]-> x -[:rel2]-> y -[:rel3]-> z, ()-[:rel1]->y, ()-[:rel1]->z 
RETURN z

推荐答案

您可以使用WHERE ALL来做到这一点，就像这样:

You can do this using WHERE ALL, like this:

START n=node(0) 
MATCH path = n -[:rel1]-> x -[:rel2]-> y -[:rel3]-> z 
WHERE ALL(n in tail(nodes(path)) WHERE ()-[:rel1]->n) 
RETURN z

tail(nodes(path))返回路径中除第一个节点外的所有节点.在您的示例中，起始节点未与rel1关系连接，因此未返回任何内容.如果您想按照文字说明进行操作，只需将尾部放下即可.

tail(nodes(path)) returns all nodes in the path except the first one. In your example, the start node was not connected with a rel1 relationship, so nothing was returned. If you want to do it like your text explains it, just drop the tail part.

这是您要找的吗?

这篇关于在路径上的所有节点上执行MATCH的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！