Question

这是一张简单的社交地图。我正在寻找节点'A的朋友们的朋友。

CREATE (pa:Person { name: "A", from: "Sweden", learn: "surfing" }),
(pb:Person { name: "B", from: "London", title: "author" }),
(pc:Person { name: "C", from: "Paris" }),
(pd:Person { name: "D", from: "Pune" }),
(pe:Person { name: "E", from: "Bangalore" }),
(pf:Person { name: "F", from: "Chicago" }),
(pg:Person { name: "G", from: "Sudney" }),
(pa)-[:KNOWS {since: 2001}]->(pb),
(pa)-[:KNOWS {rating: 5}]->(pc),
(pa)-[:KNOWS]->(pd),
(pb)-[:KNOWS]->(pd),
(pb)-[:KNOWS]->(pe),
(pb)-[:KNOWS]->(pf),
(pc)-[:KNOWS]->(pd),
(pd)-[:KNOWS]->(pa),
(pd)-[:KNOWS]->(pg)

我运行了以下查询。

MATCH (m:Person)-[:KNOWS]->(n:Person)-[:KNOWS]->(o:Person)
WHERE m.name="A"
return o;

返回A，D，E，F，G

我期待只看到E，F，G节点。

你能帮忙吗，这里有什么问题？

Answer 1

它返回A，因为A - [：知识] - ＆gt; D - [：知识] - ＆gt; A
它返回D，因为A - [：知识] - ＆gt; B - [：知识] - ＆gt; D

您需要指定m＆lt;＆gt; n＆lt;＆gt; o和m不应该知道o。

MATCH (m:Person)-[:KNOWS]->(n:Person)-[:KNOWS]->(o:Person) 
WHERE m.name="A" AND m<>o AND NOT (m)-[:KNOWS]->(o)
RETURN DISTINCT o;

请参阅：http://console.neo4j.org/r/8i2vvu

检索节点A的朋友的朋友

1 个答案: