在neo4j上，我有这样一种错误：预期的路径是集合<；路径>；？

我目前正在使用neo4j。我正试图通过neo4j研究图形数据库。为了我的学习，我需要执行一些查询。我对这个问题有个疑问：

Match path = (a:Node)-[:Rel1|Rel2*7..8]-(b:Node), (c:Node)

Where (a:Node)-[:Rel1|Rel2*7..8]-(b:Node) and (a:Node)-[:Rel1|Rel2*7..8]-(c:Node) and (b:Node)-[:Rel1|Rel2*7..8]-(c:Node)

With a as aa, b as bb, c as cc, (a:Node)-[:Rel1|Rel2*7..8]-(b:Node) as path1,
(a:Node)-[:Rel1|Rel2*7..8]-(c:Node) as path2, (c:Node)-[:Rel1|Rel2*7..8]-(b:Node) as path3

Limit 1

With a as aa, b as bb, c as cc,
path1 as path1, path2 as path2, path3 as path3, Extract(rel in relationships(path1) | rel.id) as relIDs1, Extract(rel in relationships(path2) | rel.id) as relIDs2, Extract(rel in relationships(path3) | rel.id) as relIDs3

Return path1, path2, path3

我对第二条有问题，neo4j在注释这一行时确实返回一条路径（我只需要1条路径“Limit 1”），但当我尝试执行预期路径时，却在摘录上被收集。我如何处理这个问题，我试着查看文档，但我不知道

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非常感谢

Neo4j的行为是将数据作为集合返回，即使只有一项。如果返回类型不一致，则会更加混乱，因此这确实是一件好事。只需从集合中获取第一个元素，将其存储在

Path

type变量中，然后继续