Question

start n=node(22), p=node(1) match n<-[r1:FOLLOWS]-m-[r2:HAS]->k<-[r3:CONTAIN]-p return distinct [k.name]

我试图在这里返回k值的名称和重复次数，到目前为止，我无法做到这一点。有没有快速的方法来使用密码查询？

考虑这个例子：

["Acting","Acting","Acting","Acting","Mongodb","Mongodb","Neo4j","Neo4j","Nursing"]

我正在尝试这样的事情：

[["Acting",4], ["Mongodb",2], ["Neo4j",2], ["Nursing",1]]

注意：相同的名称（属性）表示相同的节点。

Answer 1

发现它！：）

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

Answer 2

如果您想进一步过滤计数（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。很好又整洁。

    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

返回具有名称属性和重复次数的集合

2 个答案: