Question

我的Cypher查询：

MATCH p =(o:Order)-[r:seeks*2..8]->(o:Order)
WHERE o.Name="000093" AND ALL(x IN tail(nodes(p)) WHERE SINGLE(y IN    tail(nodes(p)) WHERE x=y))
RETURN extract(n IN nodes(p)| n.Name) AS OrderID, extract(u IN nodes(p)| u.UserName) AS UserName,length(p), endNode(r[0])
ORDER BY length(p)

我想避免路径中具有相同属性值的节点，如何避免它们？

["000093","000090","000096","000097","000107","000091","000089","000093"]
["yunis","gio","Anhar","Jhon","**shakilbit**","xalima","**shakilbit**","yunis"]

所以，订单0000107和000089正在使用相同的用户名shakilbit，有什么办法可以避免在同一条路径上有这样的订单，谢谢！ NEO4J ..据我所知，非常帮助社区。

Answer 1

使用APOC Procedures，您可能希望将集合作为一个集合（消除重复值）并比较大小。如果存在重复项，则集合的大小将更小。

MATCH p =(o:Order)-[r:seeks*2..8]->(o:Order)
WHERE o.Name="000093" 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
// need to make some adjustments since first and last nodes are same
WHERE size(UserName) - 1 = size(apoc.coll.toSet(tail(UserName)))
RETURN extract(n IN nodes(p)| n.Name) AS OrderID, UserName, length(p), endNode(r[0])
ORDER BY length(p)

替代方法是重复你的ALL(x in tail...) WHERE single()...谓词，但是在UserName集合上（或者在现有的ALL（）谓词中包含这个检查，尽管这可能很昂贵）。您可能想要查看每个，并查看哪个更高效。

如何避免在neo4j中多次访问具有相同属性的节点？

1 个答案: