Question

我有以下查询：

MATCH (d:Document)<-[o:OCCURS_IN]-(a:Alias)
WHERE LOWER(d.content) contains 'keyword'
WITH COUNT(o) as degree, a as a ORDER BY degree DESC LIMIT 20
MATCH (a)<-[:KNOWN_AS]-(ag:Agent)
RETURN ag;

我过滤包含关键字的所有Document节点，并按照它们连接到Document节点的频率排序前20个Alias节点。之后，将收集并返回连接到Alias节点的所有代理。

这给了我一组代理节点。除此之外，我想获得这组Agent节点中的所有关系。这意味着返回的节点集应该相同。只应添加这些节点之间的关系。

如果没有其他查询，我怎样才能实现这一目标？

Answer 1

以下是如何在原始查询返回的所有Agent节点之间获得之间的所有直接关系：

MATCH (d:Document)<-[o:OCCURS_IN]-(a:Alias) WHERE LOWER(d.content) contains 'keyword' WITH COUNT(o) as degree, a ORDER BY degree DESC LIMIT 20 MATCH (a)<-[:KNOWN_AS]-(ag:Agent) WITH COLLECT(ag) AS ags UNWIND ags AS ag1 UNIWND ags AS ag2 MATCH (ag1)-[r]->(ag2) RETURN r;

Answer 2

继续查询：

MATCH (d:Document)<-[o:OCCURS_IN]-(a:Alias)
WHERE LOWER(d.content) contains 'keyword'
WITH COUNT(o) as degree, a as a ORDER BY degree DESC LIMIT 20
MATCH (a)<-[:KNOWN_AS]-(ag:Agent)
OPTIONAL MATCH (ag)-[r]-()
RETURN ag, collect(r) as relationships

Neo4j：获取一组节点内的所有关系

2 个答案: