在neo4j上我有这样的错误：预期路径但是收集<路径>？

Question

我目前正在使用neo4j。我试图通过neo4j研究图形数据库。对于我的学习，我需要执行一些查询。我遇到了这个问题：

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

我遇到第二个问题，neo4j确实返回一条路径（我只需要1个路径“限制1”）当我评论这一行时，但当我试图做那个预期的路径但是收集时该提取物。我怎么能处理这个问题，我试着看一下我不知道的文档。

非常感谢！

Answer 1

Neo4j的行为是将数据作为Collection返回，即使只有一个项目。如果返回类型不一致会更加混乱，所以这确实是件好事。只需从集合中获取第一个元素，将其存储在Path类型变量中，然后继续。