如何仅匹配两个节点之间的一种关系 [英] how to match only one relationship between two nodes

问题描述

我正在使用neo4j graph db，它在Ruby on Rails中使用， 例如: 我在tom和jerry之间有3个关系，他们合作建造了3个房子，现在我只想匹配3个房子中的1个；如何编写查询代码? 我已经试过了:这是我的代码:

i am using neo4j graph db ,it used in Ruby on Rails , for example: i have 3 relationship between tom and jerry ,they cooperated to build 3 houses, and now i just want to match 1 of 3 ;how to write the query code???
i have tried this: this is my code:

Neo4j::Session
             .query("MATCH (s1:Scholar)<-[r:COOPERATE]-(s2:Scholar) WHERE s1.id = #{@scholar.id}  RETURN  DISTINCT r ")

结果是s1和s2之间的所有关系都建立了

and the result is all relationship between s1 and s2 were founded

我只需要s1和s2之间的一种关系(我想在我的数据库中找到所有关系，但是我只需要每2个节点之间一个关系)

i just need 1 of relationship between s1 and s2 (i want to find all relationship in my db ,but i just need one between every 2 nodes)

如何解决?

推荐答案

每个相关节点仅获得一种关系:

To get only one relationship for each related node:

MATCH (s1:Scholar)<-[r:COOPERATE]-(s2:Scholar) WHERE s1.id = {s1_id} RETURN s2, collect(r)[0] AS r

请注意，我使用的参数更安全(如果数据来自用户)，并允许Neo4j缓存您的查询.要在neo4j-core gem中使用参数:

Note that I'm using parameters which are more secure (if the data is coming from a user) and allow Neo4j to cache your query. To use parameters in the neo4j-core gem:

session.query("MATCH (s1:Scholar)<-[r:COOPERATE]-(s2:Scholar) WHERE s1.id = {s1_id} RETURN s2, collect(r)[0]", s1_id: @scholar.id).map(&:r)

还有一个查询构建API，如下所示:

There is also a query building API which would look like this:

session.query
  .match('(s1:Scholar)<-[r:COOPERATE]-(s2:Scholar)')
  .where(s1: {id: @scholar.id})
  .return('s2, collect(r)[0] AS r')
  .map(&:r)

请注意，这将为您使用参数.

Note that this will use parameters for you.

此外，如果您使用neo4j gem中的ActiveNode模块(而不是进行原始查询)，则可以执行类似的操作；

Also if you use the ActiveNode module from the neo4j gem (as opposed to making raw queries) you could do something like;

class Scholar
  include Neo4j::ActiveNode
  id_property :id

  has_many :in, :cooporates_with, type: :COOPERATE, model_class: :Scholar
end

Scholar.find(@scholar.id).cooporates_with(:s2, :r).return('s2, collect(r)[0] AS r').map(&:r)

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