查询路径时如何在neo4j中更快？

Question

大约有1000个节点，3600个关系 和许多人建立关系。 有查询sql：

match p=(:Devices{name:"123.123.123.208"})-[r:Cost*1..7]->(:Devices{name:"123.123.123.20"}) 
with p,reduce(x=0,a in relationships(p)|x+a.Cost) as value 
order by value desc 
limit 1 
return p;

个人资料：

Answer 1

为避免遍历循环，您可能需要利用APOC Procedures，特别是path expansion procs：

MATCH (start:Devices{name:"123.123.123.208"}), (end:Devices{name:"123.123.123.20"})
CALL apoc.path.expandConfig(start, {terminatorNodes:[end], relationshipFilter:'Cost>', minLevel:1, maxLevel:7, uniqueness:'NODE_PATH'}) YIELD path
WITH path, reduce(x=0,a in relationships(path)|x+a.Cost) as value 
ORDER BY value DESC 
LIMIT 1 
RETURN path;

NODE_PATH唯一性确保每个路径不能多次使用一个节点。