Question

如何返回交错的节点和关系？使用Matrix电影数据库查询

    MATCH p=(a1:Actor {name:"Keanu Reeves"})-[r *0..5]-(a2:Actor {name: "Carrie-Anne Moss"})
    return [n in nodes(p)|coalesce(n.title,n.name)], [rel in relationships(p)|type(rel)]

返回两列，一列包含节点，另一列包含关系

    Keanu Reeves, The Matrix, Laurence Fishburne, The Matrix Reloaded, Carrie-Anne Moss | ACTS_IN, ACTS_IN, ACTS_IN, ACTS_IN
    ...

但我想要

    Keanu Reeves, ACTS_IN, The Matrix, ACTS_IN, Laurence Fishburne, ACTS_IN, The Matrix Reloaded, ACTS_IN, Carrie-Anne Moss
    ...

Answer 1

这曾经更容易，但是当他们在Cypher中制作Paths not Collections时，他们打破了2.0-RC1中的“简单方法”。

match p= shortestPath((kevin)-[:ACTED_IN*]-(charlize))
where kevin.name="Kevin Bacon"
and charlize.name="Charlize Theron"
with nodes(p) as ns, rels(p) as rs, range(0,length(nodes(p))+length(rels(p))-1) as idx
return [i in idx | case i % 2 = 0 when true then coalesce((ns[i/2]).name, (ns[i/2]).title) else type(rs[i/2]) end];

旧方法是：

match p= shortestPath((kevin)-[:ACTED_IN*]-(charlize))
where kevin.name="Kevin Bacon"
and charlize.name="Charlize Theron"
return [x in p | coalesce(x.name,x.title, type(x))]

这一变化带来的好处是Paths现在更加安全，尽管我很同意它们同意它们。这种查询的真实用例很少见。

Answer 2

这是另一种解决方案：

MATCH (kevin:Person {name="Kevin Bacon"}), (charlize:Person {name:"Charlize Theron"})
MATCH p= shortestPath( (kevin)-[:ACTED_IN*]-(charlize) )

WITH nodes(p)+rels(p) AS c, length(p) AS l 
RETURN reduce(r=[], x IN range(0,l) | r + (c[x]).name + type(c[l+x+1]))

将路径重新聚合到集合中，然后使用路径长度作为偏移量来访问下半部分。

Reduce用于“展平”集合，如果你不需要，它也可以。

MATCH (kevin:Person {name="Kevin Bacon"}), (charlize:Person {name:"Charlize Theron"})
MATCH p= shortestPath( (kevin)-[:ACTED_IN*]-(charlize) )

WITH nodes(p)+rels(p) AS c, length(p) AS l 
RETURN [x IN range(0,l) | [c[x]).name + type(c[l+x+1])]]

neo4j cypher交织节点和关系

2 个答案: