如何解决Neo4j Cypher:以 id 作为字典返回节点
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","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
是键?
[
{
"114s09A.1": {
"name": "114s09A.1","updated_n4j": "2020-12-21T09:56:11.256000000Z"
}
},{
"114s09A.2": {
"name": "114s09A.2","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.2","properties":{
"name": "114s09A.2","updated_n4j": "2020-12-21T09:56:11.257000000Z"
}
}
]
解决方法
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{.*})
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