如何避免重复返回不同的节点，并使用neo4j来建立关系 [英] How to avoid duplications return distinct nodes and the relation ship using neo4j

问题描述

我想返回给定的node-id相关节点及其关系道具

I would like to return for a given node-id related nodes and their relationships props

例如: ->使用属性时间戳定义双向关系

For example: -> defines a bi direction relationship with property timestamp

1234->777
777->1234
1234->999
999->1234
1234->888
888->1234

1234,777,888,999是节点ID

1234,777,888,999 are node-ids

当我执行此操作时:

    final PreparedStatement ps = conn.prepareStatement("start a = node(1234) match (a)-[k:nearby*]->(b) where a<>b return DISTINCT b, k");        
    ResultSet rs = ps.executeQuery();
    while (rs.next()) {
         Map result = (Map<String, Object>) rs.getObject("b");     
        System.out.println(result.toString());
    }

} catch (SQLException e) {
    e.printStackTrace();
    logger.error("Error returning userId=" + userIdInput, e);
}
return null;

}

我得到:

{userId=777}
{userId=999}
{userId=888}
{userId=888}
{userId=999}
{userId=999}
{userId=777}
{userId=888}
{userId=888}
{userId=777}
{userId=888}
{userId=777}
{userId=999}
{userId=999}
{userId=777}

1. 我如何仅获得不同的结果(777,888,999)
2. 如何检索1234与dest节点的关系道具?我希望获得在每个关系上定义的时间戳道具

谢谢你， 射线.

推荐答案

我不确定您使用的是哪种语言，因此我将重点介绍Cypher.首先，我将在ID(a)上的WHERE替换为MATCH的START查询:

I'm not sure what language you're using so I'll focus on the Cypher. Firstly I would replace the START query with a MATCH with a WHERE on ID(a):

MATCH (a)-[k:nearby*]->(b)
WHERE ID(a) = 1234 AND a<>b
RETURN DISTINCT b, k

第二，我很确定您不需要a<>b，因为Cypher路径不会在相同的节点上循环:

Secondly I'm pretty sure you don't need the a<>b because Cypher paths won't loop back on the same nodes:

MATCH (a)-[k:nearby*]->(b)
WHERE ID(a) = 1234
RETURN DISTINCT b, k

最后，对于您的问题，我怀疑您重复的原因是因为您有多个关系.如果是这样，您可以返回结果节点和关系数组，如下所示:

Lastly, and to your question, I suspect the reason that you're getting duplicates is because you have multiple relationships. If so you can return the result node and an array of the relationships like so:

MATCH (a)-[k:nearby*]->(b)
WHERE ID(a) = 1234
RETURN collect(b), k

那应该返回您的节点/关系对象(两个都有属性).根据您的语言/图书馆，您可能会获得地图，或者可能会得到包装数据的对象

That should return you node/relationship objects (with properties on both). Depending on your language/library you might get Maps or you might get objects wrapping the data

如果您的库没有为您返回关系的开始/结束节点，则可以执行以下操作:

If your library doesn't return the start/end nodes for relationships for you, you can do something like this:

MATCH (a)-[k:nearby*]->(b)
WHERE ID(a) = 1234
RETURN collect({rel: b, startnode: startnode(b), endnode: endnode(b)}), k

希望有帮助！

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