neo4j - 如何知道节点 x 是否具有节点 z 的每个实例，其中它与另一个节点 y 有很多关系，而另一个节点 y 与另一个节点 z 有一个关系？

Question

所以我有一个节点:Customer，它有很多 node :Order，并且与node:Order有 1 个关系，这里:Shipper调用:SHIP_VIA的是层次结构：

(customer:Customer)-[:PURCHASED]->(order)-[:SHIP_VIA]->(shipper:Shipper)

现在只有 3 个托运人，订单指向他们，一个客户会有很多订单。所以我的问题是如何获得与所有托运人一起发货的所有客户？请注意，每个托运人shipperID内部都有一个属性。这是我的代码，但除了一个“存在的地方”之外它不起作用

match (customer:Customer)-[:PURCHASED]->(order)-[:SHIP_VIA]->(shipper:Shipper) 

WHERE exists((customer)-[:PURCHASED]->(order)-[:SHIP_VIA]->(:Shipper 

{shipperID:1})) and exists((customer)-[:PURCHASED]->(order)-[:SHIP_VIA]->

(:Shipper {shipperID:2})) and exists((customer)-[:PURCHASED]->(order)-

[:SHIP_VIA]->(:Shipper {shipperID:3})) return  customer, order , shipper;

标签： neo4jcypher

Answer 1

使用collect()和all()函数尝试此解决方案：

// match all shippers
match (shipper:Shipper)
// create a list containing all shippers and pass to the next context
with collect(shipper) as shippers

// match all customers which has purchased an order with each shipper
match (customer:Customer)
where all (shipper in shippers where (customer)-[:PURCHASED]->(:Order)-[:SHIP_VIA]->(shipper))

// return customers
return customer