Question

我从这个答案中得到了帮助它运行正常，但对于孤立节点而言并不意味着没有关系的单个节点（子节点）。在这种情况下我得到的甚至不是那个节点。请帮助我，我是neo4j的初学者，这将是非常有利的

OPTIONAL MATCH path = (x)-[*0..100]->()       
WHERE ID(x) = 65
UNWIND nodes(path) as node
UNWIND rels(path) as rel

WITH collect(distinct node) as nodes,collect(distinct rel) as rels
// WITH apoc.coll.flatten(collect(nodes(path))) as nodes, apoc.coll.flatten(collect(relationships(path))) as rels
WITH apoc.coll.toSet([n in nodes WHERE n is not null 
            | { id: id(n),label: labels(n),type:"",metadata: properties(n)  } ]) as nodes,
     apoc.coll.toSet([r in rels WHERE r is not null 
            | { id: id(r),source: id(startNode(r)),relation: type(r),target: id(endNode(r)), directed: "true"  } ]) as rels

RETURN { graph: { type:"",label: "",directed: "true",nodes: nodes,edges: rels,
         metadata:{ countNodes: size(nodes),countEdges: size(rels) } } } as graph;

由于

Answer 1

问题是：UNWIND rels(path) as rel

UNWINDing集合意味着获取该记录，并将其更改为集合中每个元素的记录，它会使记录倍增。当集合为空（孤立节点没有关系）时，它的乘数为零......它会清空带有空集合的记录。

您可以使用CASE语句将集合替换为null而不是空集合（当您再次收集时，null将被清除）。当你UNWIND时，它会保留记录。

UNWIND case when size(rels(path)) = 0 then [null] else rels(path) end  as rel

没有在neo4j中获得孤立节点

1 个答案: