Neo4j密码查询：获取最后N个元素 [英] Neo4j cypher query: get last N elements

问题描述

我有一个图表，其中包含用户之间的关系，显示哪个用户访问了另一个用户的个人资料，以及何时：

I have a graph that contains a relationship between users, showing what user has visited another's profile, and when:

(visitor:User)-[:VISITED]->(visitee:User)

但是我不从时间开始起，我们就不想存储每次访问。我只想要一个用户访问过的最后X个用户。因此，在添加新关系之前，我必须删除最旧的关系，但是我不知道如何删除它。我只能按日期排序：

But I don't want to store every visit since the beggining of time. I only want the last X users that a user has visited. So before adding a new relationship I must delete the oldest one, but I don't know how to delete it. I can only get a list ordered by date:

MATCH (visitor:User)-[r:VISITED]->(User)
WHERE visitor.user_id = %s
RETURN r
ORDER BY r.date

我需要删除此列表中的第一个关系。我该怎么办？

What I need is to delete the first relationship in this list. How can I do that?

推荐答案

最简单的方法是使用 ORDER BY 和 LIMIT 在 DELETE 之前。即：

The simplest would be to use an ORDER BY and a LIMIT before the DELETE. I.e.:

MATCH (visitor:User)-[r:VISITED]->(User)
WHERE visitor.user_id = %s
WITH r ORDER BY r.date LIMIT 1
DELETE r

一种更有效的机制可能是保存访问的链接列表（此处有一些讨论： http://docs.neo4j.org/chunked/stable/cypherdoc-linked-lists.html ）

A more efficient mechanism may be to keep a linked list of visit (there's some discussion on this here: http://docs.neo4j.org/chunked/stable/cypherdoc-linked-lists.html)

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