Neo4J |关系列表上的密码汇总 [英] Neo4J | Cypher Aggregation on list of relationships

问题描述

Neo4J/Cypher的新功能，这是我的查询:

New to Neo4J/Cypher and here is my query:

MATCH (origin:BusStop)-[bus*]->(destination:BusStop)
WITH bus
WHERE origin.name =~ '(?i).*Origin.*' 
AND destination.name =~ '(?i).*Destination.*' 
AND all(rel in bus where rel.day in ['Sat'])
RETURN bus

我正在尝试获取始发地和目的地之间的所有可能的巴士.我还希望上述查询中的总票价(作为总和功能).

I'm trying to get all the possible buses between the Origin and Destination. I also want the total fare (as a sum function) in the above query.

注意:该关系具有一个称为fare SUM(bus.fare))的属性.

Note: The relationship has a property called fare SUM(bus.fare)).

推荐答案

使用 APOC程序，您可以对集合的元素求和，尽管您需要从公交集合中的每个关系中提取票价:

With APOC Procedures you can sum elements of a collection, though you'll need to extract the fare value from each relationship in the bus collection:

RETURN bus, apoc.coll.sum([rel in bus | rel.fare]) as totalFare

没有APOC程序，您将需要使用REDUCE():

Without APOC Procedures, you'll need to use REDUCE():

RETURN bus, reduce(total = 0, rel in bus | total + rel.fare) as totalFare

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