Neo4j

Question

我有一个树结构，对我的树的每个节点都有访问权限，我想在这样的嵌套对象中只获取我有权访问的节点：

{
  name: "project 1",
  Children: [
   {name: "nested child"}
  ]
}

这是我的图结构：

我想获取所有与 Marion 有 :VIEWER 关系的节点，这是我试过的查询，但我没有嵌套结构

MATCH path =(a:Account {firstName: "Marion"})-[:VIEWER]->()
WITH collect(path) AS paths
CALL apoc.convert.toTree(paths)
YIELD value
RETURN value;

我得到这个结构

{
  "firstName": "Marion",
  "viewer": [
    {
      "_type": "Project",
      "name": "project 1",
      "_id": 1,
      "type": "project"
    },
    {
      "_type": "Child",
      "name": "nested child",
      "_id": 4,
      "type": "child"
    }
  ],
  "_type": "Account",
  "_id": 29,
}

nested child 应该在 children 对象中 project1

父->子关系的深度没有限制

有谁知道这是否可行？

如果您对我的图形结构有更好的想法，我愿意进行优化

EDIT

感谢您的回答，它有效，但我想添加一些参数

不仅可以有其他类型的关系 :VIEWER 还可以有 :VIEWER :EDITOR :OWNER

我需要知道每个节点的关系Marion

我已经进行了此查询，但我不明白为什么我的结果被拆分成表格？

MATCH (c)<-[:IS_PARENT_OF*]-(p)
WHERE (p)<-[:VIEWER | EDITOR]-(a)
// collect the child nodes per project
WITH TYPE(r) as relation, p,collect(distinct c) AS children
// return a map that contains project and children names
RETURN {name: p.name, Children: [c in children | {name: c.name, access:relation}]}

RESULT

{
  "name": "project 1",
  "Children": [
    {
      "access": "EDITOR",
      "name": "Experimental design"
    }
  ]
}

{
  "name": "project 1",
  "Children": [
    {
      "access": "VIEWER",
      "name": "nested child"
    }
  ]
}

EXPECTED

{
  "name": "project 1",
  "Children": [
 {
      "access": "EDITOR",
      "name": "Experimental design"
    },
    {
      "access": "VIEWER",
      "name": "nested child"
    }
  ]
}

Answer 1

根据您的节点可视化，我假设项目节点与子节点具有不同的标签。

我会尝试以下方法：

// get the top-level project
MATCH (a:Account {firstName: "Marion"})-[:VIEWER]->(p:Project)
// get the child nodes
MATCH (p)-[:IS_PARENT_OF*]->(child)
WHERE (child)<-[:VIEWER]-(a)
// collect the child nodes per project
WITH p, collect(distinct child) AS children
// return a map that contains project and children names
RETURN {name: p.name, Children: [c in children | {name: c.name}]}

编辑：

添加了 Hakan 的回答

// get the children
MATCH (a:Account {firstName: "Marion"})-[:VIEWER]->(child:Child) 
// get the parent project that is also viewable 
MATCH (p)-[:IS_PARENT_OF*]->(child) 
WHERE (p)<-[:VIEWER]-(a) 
// collect the child nodes per project 
WITH p, collect(distinct child) AS children 
// return a map that contains project and children names 
RETURN {name: p.name, Children: [c in children | {name: c.name}]}

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