我目前正在使用console.neo4j.org上的示例数据编写一个输出分层JSON的查询.
使用创建示例数据
create (Neo:Crew {name:'Neo'}), (Morpheus:Crew {name: 'Morpheus'}), (Trinity:Crew {name: 'Trinity'}), (Cypher:Crew:Matrix {name: 'Cypher'}), (Smith:Matrix {name: 'Agent Smith'}), (Architect:Matrix {name:'The Architect'}),
(Neo)-[:KNOWS]->(Morpheus), (Neo)-[:LOVES]->(Trinity), (Morpheus)-[:KNOWS]->(Trinity),
(Morpheus)-[:KNOWS]->(Cypher), (Cypher)-[:KNOWS]->(Smith), (Smith)-[:CODED_BY]->(Architect)
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理想的输出如下
name:"Neo"
children: [
{
name: "Morpheus",
children: [
{name: "Trinity", children: []}
{name: "Cypher", children: [
{name: "Agent Smith", children: []}
]}
]
}
]
}
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现在,我正在使用以下查询
MATCH p =(:Crew { name: "Neo" })-[q:KNOWS*0..]-m
RETURN extract(n IN nodes(p)| n)
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得到这个
[(0:Crew {name:"Neo"})]
[(0:Crew {name:"Neo"}), (1:Crew {name:"Morpheus"})]
[(0:Crew {name:"Neo"}), (1:Crew {name:"Morpheus"}), (2:Crew {name:"Trinity"})]
[(0:Crew {name:"Neo"}), (1:Crew {name:"Morpheus"}), (3:Crew:Matrix {name:"Cypher"})]
[(0:Crew {name:"Neo"}), (1:Crew {name:"Morpheus"}), (3:Crew:Matrix {name:"Cypher"}), (4:Matrix {name:"Agent Smith"})]
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有任何提示来解决这个问题吗?谢谢
cyb*_*sam 10
在neo4j 3.x中,在neo4j服务器上安装APOC插件后,可以调用该apoc.convert.toTree过程来生成类似的结果.
例如:
MATCH p=(n:Crew {name:'Neo'})-[:KNOWS*]->(m)
WITH COLLECT(p) AS ps
CALL apoc.convert.toTree(ps) yield value
RETURN value;
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...将返回如下所示的结果行:
{
"_id": 127,
"_type": "Crew",
"name": "Neo",
"knows": [
{
"_id": 128,
"_type": "Crew",
"name": "Morpheus",
"knows": [
{
"_id": 129,
"_type": "Crew",
"name": "Trinity"
},
{
"_id": 130,
"_type": "Crew:Matrix",
"name": "Cypher",
"knows": [
{
"_id": 131,
"_type": "Matrix",
"name": "Agent Smith"
}
]
}
]
}
]
}
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