我是neo4j的新手 .
我创建了这个例子:
// (Node2)
// / \
// Rel2 / \ Rel3
// / \
// (Node1) ------> (Node3)
// Rel1
//
// create nodes
CREATE
(n1:Node1{id:1,rights:["admin","user"]}),
(n2:Node2{id:2,rights:["admin","user"]}),
(n3:Node3{id:3,rights:["admin","user"]})
// create relationships
CREATE (n1)-[r1:Rel1{id:11,rights:["admin"]}]->(n3)
CREATE (n1)-[r2:Rel2{id:12,rights:["admin","user"]}]->(n2)
CREATE (n2)-[r3:Rel3{id:13,rights:["admin","user"]}]->(n3)
RETURN n1,n2,n3,r1,r2,r3
每个节点和关系都有一个属性数组,其中包含一些权限值 . 我想创建一个查询,它给出了具有指定属性的两个节点之间的(最短)路径 . 例:
用户拥有权限'admin',路径应为:
(节点1) - [REL1] - >(节点3)
(每个节点和相关性在'rights'属性中都有'admin'字符串 . )
如果用户拥有正确的“用户”,则路径应为:
(节点1) - [REL2] - (节点2) - [REL3] - (节点3)
(因为Rel1关系没有't have the ' user ' String in the ' rights'属性 . )
起初我尝试了这个有效的查询:
WITH ["admin","otherRight"] AS usersRights
MATCH path=allShortestPaths((n1:Node1{id:1})-[*..4]-(n2:Node3{id:3}))
WITH *, relationships(path) AS rels, nodes(path) as nodes
WHERE ANY (rel IN rels WHERE ANY(r IN rel.rights WHERE r IN usersRights))
AND ANY (node IN nodes WHERE ANY(r IN node.rights WHERE r IN usersRights))
RETURN path
然后我将'admin'替换为'user',但是这个查询不起作用(没有rusults):
WITH ["user","otherRight"] AS usersRights
MATCH ... // same as above
...
RETURN path
以下查询匹配我需要的结果,但在这种情况下,我将不得不进行多次查询以获得深度n的所需结果(最短路径) .
WITH ["user","otherRight"] AS usersRights
MATCH path=(n1:Node1{id:1})-[r1]-(n2)-[r2]-(n3:Node3{id:3})
WHERE ANY(r IN n1.rights WHERE r IN usersRights)
AND ANY(r IN r1.rights WHERE r IN usersRights)
AND ANY(r IN n2.rights WHERE r IN usersRights)
AND ANY(r IN r2.rights WHERE r IN usersRights)
AND ANY(r IN n3.rights WHERE r IN usersRights)
RETURN path
是否有可能创建一个cypher查询,让我得到令人反感的结果?
谢谢 .
1 回答
您的第一个查询具有分隔
MATCH
和WHERE
子句的WITH
子句,这会阻止WHERE
子句影响allshortestpath()
函数的行为 . 因此,该函数只返回最短路径而不测试节点和关系 .试试这个,相反: