在 Cypher 中多次遍历关系

发布于 2025-01-10 07:36:41 字数 1078 浏览 0 评论 0原文

我有一张机场图、机场之间的路线以及承运该机场的航空公司的图表。我将路线创建为单独的节点，而不仅仅是一种关系，以便我可以将每个路线与航空公司和其他节点连接起来。

每个Route 节点与始发机场有一个IS_FROM 关系，与目的地机场有一个IS_TO 关系。它还与其航空公司有 IS_BY 关系：

我正在尝试遍历这棵树n次，以获取两个机场之间的路线。例如，如果n = 3，我想获取从LAX到LHR的所有路线，有3个或更少的连接。

基本上，我的结果将是以下各项的并集：无中转机场：

MATCH (a1:Airport {iata : 'LAX'})<-[:IS_FROM]-(r:Route)-[:IS_TO]->(a2:Airport {iata : 'LHR'}), (r)-[:IS_BY]->(ai:Airline) return a1 , r , a2 , ai;

1 中转机场：

MATCH (a1:Airport {iata : 'LAX'})<-[:IS_FROM]-(r:Route)-[:IS_TO]->(a2:Airport)<-[IS_FROM]-(r2:Route)-[:IS_TO]->(a3:Airport {iata: 'LHR'}), (r2)-[:IS_BY]->(ai:Airline) return a1 , r , a2 , a3 , r2 , ai;

等等。

因此查询应该动态遍历 (:Airport)<-[:IS_FROM]-(:Route)-[:IS_TO]->(:Airport) 模式 n 次，并返回节点（我更感兴趣的是返回连接到这些路线的航空公司。

原文

I have a graph of Airports, Routes between them and Airlines that carry it. I created routes as separate nodes, rather than just a relationship, so that I can connect each with an Airline, and other nodes.

Each Route node has an IS_FROM relationship with the origin airport and an IS_TO relationship with the destination. It also has an IS_BY relationship with its airline:

I am trying to traverse this tree, n times, for routes between two airports. For example, if n = 3, I want to get all the routes, that will lead from LAX to LHR, with 3 or fewer connections.

So basically, my result would be a union of the following:
No Connecting Airports:

MATCH (a1:Airport {iata : 'LAX'})<-[:IS_FROM]-(r:Route)-[:IS_TO]->(a2:Airport {iata : 'LHR'}), (r)-[:IS_BY]->(ai:Airline) return a1 , r , a2 , ai;

1 Connecting airports:

MATCH (a1:Airport {iata : 'LAX'})<-[:IS_FROM]-(r:Route)-[:IS_TO]->(a2:Airport)<-[IS_FROM]-(r2:Route)-[:IS_TO]->(a3:Airport {iata: 'LHR'}), (r2)-[:IS_BY]->(ai:Airline) return a1 , r , a2 , a3 , r2 , ai;

and so on.

So the query should dynamically traverse the (:Airport)<-[:IS_FROM]-(:Route)-[:IS_TO]->(:Airport) pattern n times, and return the nodes (I am more interested in returning the Airlines that connect to those routes.

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彻夜缠绵 2025-01-17 07:36:41

您可以首先提取距离 3 跳或更少的机场之间的所有路径，然后使用 OPTIONAL MATCH 来查看路径中的哪些节点是路线，以及哪些航空公司提供这些路线。

MATCH path = (:Airport {iata:$start_airport})-[*2..6]-(:Airport {iata:$end_airport})
WITH path,nodes(path) as path_airports_and_connecting_routes
UNWIND path_airports_and_connecting_routes as node
OPTIONAL MATCH (node)-[:IS_BY]-(airline:Airline)
WITH collect(node) as airports_and_routes,airline
RETURN airports_and_routes + [airline]

注意：可变长度路径不允许传递参数，因此您不能执行类似 [*2..2*n] 的操作。

You can first extract all the paths between Airports that are 3 or less hops away, and then use OPTIONAL MATCH to see which of the nodes in the path are Routes, and which Airlines are offering them.

MATCH path = (:Airport {iata:$start_airport})-[*2..6]-(:Airport {iata:$end_airport})
WITH path,nodes(path) as path_airports_and_connecting_routes
UNWIND path_airports_and_connecting_routes as node
OPTIONAL MATCH (node)-[:IS_BY]-(airline:Airline)
WITH collect(node) as airports_and_routes,airline
RETURN airports_and_routes + [airline]

Caveat: Variable-length paths don't allow passing parameters, so you can't do something like [*2..2*n].

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煮酒 2025-01-17 07:36:41

我不知道你的问题我是否答对了。对我来说你的问题可以这样解决：

MATCH (a1:Airport {iata : 'LAX'})<-[r1:IS_FROM]-(r:Route)-[r2:IS_TO]->(a2:Airport{iata : 'LHR'})
OPTIONAL MATCH (r)-[r3:IS_BY]->(ai:Airline)
RETURN a1,r1,r,r2,a2,r3,ai

I don't know if i got your question right. To me your problem could be solved this way:

MATCH (a1:Airport {iata : 'LAX'})<-[r1:IS_FROM]-(r:Route)-[r2:IS_TO]->(a2:Airport{iata : 'LHR'})
OPTIONAL MATCH (r)-[r3:IS_BY]->(ai:Airline)
RETURN a1,r1,r,r2,a2,r3,ai

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