Question

我有以下图表 graph layout

如您所见，该图表具有以下关系：

(u::4)-[ADDED_RESOURCE]->(:resource)<-[ADDED_RESOURCE]-(u::3) \\ u::4, u::3 are the ids of the nodes。
(u::4)-[UNLINK]->(u::3)

我使用APOC像这样遍历图表：

MATCH (u:user {id:"u::1"}
CALL apoc.path.expandConfig(u,{minLevel:1,maxLevel:6,bfs:true,uniqueness:"NODE_PATH",labelFilter:">resource"}) YIELD path
with u, path, filter(n in nodes(path) where n:resource) as resources
unwind resources as resource
MATCH (rus:user)-[]->(resource)
RETURN distinct rus.id

通过它的相关资源返回与节点u::X相关的所有u::1个节点。

由于u::4和u::3取消关联，我希望遍历忽略该连接，而不返回与u::3相关的子图。因此，它不应返回u::4, u::3, u::2, u::5，而应仅返回u::4。

有没有办法告诉APOC在遍历时忽略它们之间有某种关系的节点？

Answer 1

我不认为apoc.path.expandConfig会允许您忽略关系类型列表，但它会遵循正面表达的关系类型。它可以选择使用<，>来计算订单。

MATCH (u:user {id:"u::1"}
CALL apoc.path.expandConfig(u
  {
    minLevel:1,
    maxLevel:6,
    bfs:true,
    uniqueness:"NODE_PATH",
    labelFilter:">resource",

    // add relationship filter to folow only relationships that are included
    relationshipFilter: 'ADDED_RESOURCE|OTHER_TYPE|...'
}) YIELD path
with u, path, filter(n in nodes(path) where n:resource) as resources
UNWIND resources as resource
MATCH (rus:user)-[]->(resource)
RETURN distinct rus.id

Cypher - 从结果中移除子图

1 个答案: