如何獲取路徑中的節點屬性的總值（Neo4j）？

Question

MATCH p =(o:Order)-[r:seeks*2..8]->(o:Order) 
WHERE o.Name={orderID} AND ALL(x IN tail(nodes(p)) WHERE 
SINGLE(y IN tail(nodes(p)) WHERE x=y)) 
WITH p, o, r, extract(u IN nodes(p)| u.UserName) AS UserName 
WHERE size(UserName) - 1 = size(apoc.coll.toSet(tail(UserName))) 
RETURN extract(n IN nodes(p)| n.Name) AS OrderID, UserName, length(p) as 
Participants, endNode(r[0]) as AvailableOffers, ***reduce(s = 0, x IN 
extract(b IN nodes(p)| coalesce(b.userReputation, 0)) | s + x) AS UserRep*** 
ORDER BY length(p)""" 

我不能得到總的使用減少功能中的一個節點的屬性值...

Answer 1

我無法重現您的查詢，但我想我找到了你的問題是使用既減少又提取。試試這個

MATCH p =(o:Order)-[r:seeks*2..8]->(o:Order) 
WHERE o.Name={orderID} AND ALL(x IN tail(nodes(p)) WHERE 
SINGLE(y IN tail(nodes(p)) WHERE x=y)) 
WITH p, o, r, extract(u IN nodes(p)| u.UserName) AS UserName 
WHERE size(UserName) - 1 = size(apoc.coll.toSet(tail(UserName))) 
RETURN extract(n IN nodes(p)| n.Name) AS OrderID, UserName, length(p) as 
Participants, endNode(r[0]) as AvailableOffers, ***reduce(s = 0, x IN 
nodes(p)| s + coalesce(x.userReputation, 0)) AS UserRep*** 
ORDER BY length(p)"""