目前正在分析两个节点之间的不同路径。为此,我需要某种方法来唯一地识别Neo4J返回的路径中的路径。
测试用例:
#MATCH (n:City) DETACH DELETE n;
CREATE (n:City { name: 'Rome' });
CREATE (n:City { name: 'London' });
CREATE (n:City { name: 'Berlin' });
CREATE (n:City { name: 'Madrid' });
MATCH (n:City),(m:City)
WHERE n.name = 'London' AND m.name = 'Madrid'
CREATE (n)-[r:Connected]->(m) RETURN r;
MATCH (n:City),(m:City)
WHERE n.name = 'Madrid' AND m.name = 'Berlin'
CREATE (n)-[r:Connected]->(m) RETURN r;
MATCH (n:City),(m:City)
WHERE n.name = 'Berlin' AND m.name = 'Rome'
CREATE (n)-[r:Connected]->(m) RETURN r;
MATCH (n:City),(m:City)
WHERE n.name = 'London' AND m.name = 'Berlin'
CREATE (n)-[r:Connected]->(m) RETURN r;
MATCH (n:City),(m:City) WHERE n.name = 'London' AND m.name = 'Rome'
CREATE (n)-[r:Express]->(m) RETURN r;
以下查询正常
MATCH (n:City),(m:City)
WHERE n.name = 'London' AND m.name = 'Rome'
with n,m
match p= (n)-[*1..]->(m)
return p;
+---------------------------------------------------------------------------------------------------------------------------------------------+
| p |
+---------------------------------------------------------------------------------------------------------------------------------------------+
| (:City {name: "London"})-[:Connected]->(:City {name: "Madrid"})<-[:Connected]-(:City {name: "Berlin"})-[:Connected]->(:City {name: "Rome"}) |
| (:City {name: "London"})-[:Connected]->(:City {name: "Berlin"})<-[:Connected]-(:City {name: "Rome"}) |
| (:City {name: "London"})-[:Express]->(:City {name: "Rome"}) |
+---------------------------------------------------------------------------------------------------------------------------------------------+
为了获取访问过的节点,我使用了查询:
MATCH (n:City),(m:City)
WHERE n.name = 'London' AND m.name = 'Rome'
with n,m
match p= (n)-[*1..]->(m)
unwind nodes(p) as n1 return n1;
+--------------------------+
| n1 |
+--------------------------+
| (:City {name: "London"}) |
| (:City {name: "Madrid"}) |
| (:City {name: "Berlin"}) |
| (:City {name: "Rome"}) |
| (:City {name: "London"}) |
| (:City {name: "Berlin"}) |
| (:City {name: "Rome"}) |
| (:City {name: "London"}) |
| (:City {name: "Rome"}) |
+--------------------------+
但是,我需要一个属性来确定获取节点的路径,例如
+--------------------------+---------+
| n1 | pathid |
+--------------------------+---------+
| (:City {name: "London"}) | 1 |
| (:City {name: "Madrid"}) | 1 |
| (:City {name: "Berlin"}) | 1 |
| (:City {name: "Rome"}) | 1 |
| (:City {name: "London"}) | 2 |
| (:City {name: "Berlin"}) | 2 |
| (:City {name: "Rome"}) | 2 |
| (:City {name: "London"}) | 3 |
| (:City {name: "Rome"}) | 3 |
+--------------------------+---------+
我尝试使用cypher foreach
和reduce
运算符但没有成功。能否请你就我如何列举路径提供一些指示?
由于 肯尼斯
答案 0 :(得分:2)
我认为这样的事情可能适合你。
MATCH (n:City),(m:City)
WHERE n.name = 'London' AND m.name = 'Rome'
WITH n,m
MATCH p= (n)-[*1..]->(m)
WITH COLLECT(p) as paths
UNWIND RANGE(0,size(paths)-1) AS idx
UNWIND NODES(paths[idx]) AS city
RETURN city, idx
如果你想把它放在一点点饮食上,你也可以重写这样的测试数据。
CREATE (n1:City { name: 'Rome' })
CREATE (n2:City { name: 'London' })
CREATE (n3:City { name: 'Berlin' })
CREATE (n4:City { name: 'Madrid' })
CREATE (n2)-[:Connected]->(n4)
CREATE (n4)-[:Connected]->(n3)
CREATE (n3)-[:Connected]->(n1)
CREATE (n2)-[:Connected]->(n3)
CREATE (n2)-[:Express]->(n1);