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Neo4j 返回具有name属性和重复时间的集合_Neo4j_Cypher - Fatal编程技术网

Neo4j 返回具有name属性和重复时间的集合

Neo4j 返回具有name属性和重复时间的集合,neo4j,cypher,Neo4j,Cypher,我试着做一些类似于: ["Acting","Acting","Acting","Acting","Mongodb","Mongodb","Neo4j","Neo4j","Nursing"] 注意:相同的名称(属性)表示相同的节点。找到它了!:) start n=node(22),p=node(1)匹配nk,如果您想要进一步过滤计数(假设您只想要计数大于5的结果,您也可以将其包装在WITH语句中。漂亮整洁 start n=node(22), p=node(1) match n<-[r1:F

我试着做一些类似于:

["Acting","Acting","Acting","Acting","Mongodb","Mongodb","Neo4j","Neo4j","Nursing"]
注意:相同的名称(属性)表示相同的节点。

找到它了!:)


start n=node(22),p=node(1)匹配nk,如果您想要进一步过滤计数(假设您只想要计数大于5的结果,您也可以将其包装在WITH语句中。漂亮整洁

start n=node(22), p=node(1) match n<-[r1:FOLLOWS]-m-[r2:HAS]->k<-[r3:CONTAIN]-p return distinct [k.name] as skill, count(k) as count
START n=节点(22),p=节点(1)
匹配nk 5
还击技能,计数
[["Acting",4], ["Mongodb",2], ["Neo4j",2], ["Nursing",1]]
start n=node(22), p=node(1) match n<-[r1:FOLLOWS]-m-[r2:HAS]->k<-[r3:CONTAIN]-p return distinct [k.name] as skill, count(k) as count
    START n=node(22), p=node(1) 
    MATCH n<-[r1:FOLLOWS]-m-[r2:HAS]->k<-[r3:CONTAIN]-p 
    WITH distinct [k.name] as skill, count(k) as count
    WHERE count > 5
    RETURN skill, count