时间戳

首页 > 解决方案 > Neo4j Cypher：返回具有 id 作为字典的节点

neo4j - Neo4j Cypher：返回具有 id 作为字典的节点

问题描述

我有具有这些属性的节点：

MATCH (n:A) RETURN n

[
  {
    "name": "114s09A.1",
    "_id": "114s09A.1",
    "id": "114s09A.1",
    "created_n4j": "2020-12-21T09:56:11.256000000Z",
    "type": "A",
    "updated_n4j": "2020-12-21T09:56:11.256000000Z"
  },
  {
    "name": "114s09A.2",
    "_id": "114s09A.2",
    "id": "114s09A.2",
    "created_n4j": "2020-12-21T09:56:11.257000000Z",
    "type": "A",
    "updated_n4j": "2020-12-21T09:56:11.257000000Z"
  }
]

有没有办法构建密码查询，以便将结果塑造成字典，其中id的关键是？

[
  {
    "114s09A.1": {
      "name": "114s09A.1",
      "id": "114s09A.1",
      "created_n4j": "2020-12-21T09:56:11.256000000Z",
      "type": "A",
      "updated_n4j": "2020-12-21T09:56:11.256000000Z"
    }
  },
  {
    "114s09A.2": {
      "name": "114s09A.2",
      "id": "114s09A.2",
      "created_n4j": "2020-12-21T09:56:11.257000000Z",
      "type": "A",
      "updated_n4j": "2020-12-21T09:56:11.257000000Z"
    }
  }
]

到目前为止，我最接近的是：

MATCH (n:A) RETURN n._id AS _id, properties(n) AS properties

[
  {
    "_id":"114s09A.1",
    "properties":{
      "name": "114s09A.1",
      "id": "114s09A.1",
      "created_n4j": "2020-12-21T09:56:11.256000000Z",
      "type": "A",
      "updated_n4j": "2020-12-21T09:56:11.256000000Z"
    }
  },
  {
    "_id":"114s09A.2",
    "properties":{
      "name": "114s09A.2",
      "id": "114s09A.2",
      "created_n4j": "2020-12-21T09:56:11.257000000Z",
      "type": "A",
      "updated_n4j": "2020-12-21T09:56:11.257000000Z"
    }
  }
]

标签： neo4jcypher

解决方案

使用默认的 Cypher 语法是不可能的，但是如果你安装了 apoc 库，你可以这样做：

MATCH (n:A)
RETURN apoc.map.setKey({}, n.id, n{.*})

推荐阅读