neo4j - 在单个 Cypher 查询中返回两个聚合？

Question

我一直在与 Cypher 进行一些斗争，以求两个值的 SUM 并找出差异。我有这两个查询，它们找到一个节点的发送总数和接收总数:

START addr = node(5)
MATCH addr <- [:owns] - owner - [to:transfers] -> receiver
RETURN SUM(to.value) AS Total_Sent

START addr = node(5)
MATCH addr <- [:owns] - owner <- [from:transfers] - sender
RETURN SUM(from.value) AS Total_Received

基本上我的问题是 - 如何组合这两个单独的查询以便我可以区分 Total_Sent 和 Total_Received？我已经尝试了多个起点:

START sendAddr = node(5), receivedAddr = node(5)
MATCH sendAddr <- [:owns] - sendOwner - [to:transfers] -> receiver, receivedAddr <- [:owns] - receiveOwner <- [from:transfers] - sender
RETURN SUM(to.value) AS Total_Sent, SUM(from.value) AS Total_Received, SUM(to.value) - SUM(from.value) AS Balance

但是 Total_Received 为空!对我来说，这看起来是一个非常简单的用例 - 我到底做错了什么？

Answer 1

您不能通过像那样将两个查询拼凑在一起来组合两个查询。 :)

为了解决这个问题，我建议您使用 WITH 将您的查询分成两部分，如下所示:

START addr = node(5)
MATCH addr <- [:owns] - owner - [to:transfers] -> receiver
WITH addr, SUM(to.value) AS Total_Sent

MATCH addr <- [:owns] - owner <- [from:transfers] - sender
WITH SUM(from.value) AS Total_Received, Total_Sent

RETURN Total_Received, Total_Sent, Total_Received - Total_Sent as Total_Balance

HTH,

安德烈斯