neo4j中的双向递归

Question

无法在线找到答案。 我想对蛋白质相互作用 map 进行上游和下游递归查询。如果用户输入蛋白质（蛋白质“C”）并且深度 N=2，我想在相互作用 map 和调节中返回 2 个上游和 2 个下游蛋白质。 但是，如果它是上游，则 MATCH 右侧的蛋白质“b”需要在返回表中排在第一位，如果它是下游方向，则匹配左侧的蛋白质“a”需要在返回表中排在第一位。 我怎样才能做到这一点？

例如，这是双向的，但一半的行在第 1 列和第 3 列中的顺序是错误的。

MATCH p = (a:Protein { name:'C' })<-[:REGULATES*1..2]->(b:Protein) 
WITH *, relationships(p) as r
RETURN nodes(p)[length(p)-1].name AS Protein1, r[length(p)-1] as Regulates, b.name AS Protein2

我只能通过两次调用和切换顺序或 RETURN 列来获得我想要的东西。

MATCH p = (a:Protein { name:'C' })-[:REGULATES*1..2]->(b:Protein) 
WITH *, relationships(p) as r
RETURN nodes(p)[length(p)-1].name AS Protein1, r[length(p)-1] as Regulates, length(p), b.name AS Protein2

MATCH p = (a:Protein { name:'C' })<-[:REGULATES*1..2]-(b:Protein) 
WITH *, relationships(p) as r
RETURN b.name AS Protein1, r[length(p)-1] as Regulates, nodes(p)[length(p)-1].name AS Protein2

Answer 1

使用函数startNode和endNode 弄清楚了。 last() 和 head() 函数也很方便。

MATCH p = (n:Protein { name:'C' })<-[:REGULATES*1..3]->(b:Protein) 
WITH *, relationships(p) as rs
RETURN startNode(last(rs)).name as Protein1, last(rs).direction as Regulates, endNode(last(rs)).name as Protein2, length(p)