neo4j中的雙向遞歸

Question

無法在線找到答案。 我想對蛋白質相互作用 map 進行上游和下游遞歸查詢。如果用戶輸入蛋白質（蛋白質“C”）並且深度 N=2，我想在相互作用 map 和調節中返回 2 個上游和 2 個下游蛋白質。 但是，如果它是上游，則 MATCH 右側的蛋白質“b”需要在返回表中排在第一位，如果它是下游方向，則匹配左側的蛋白質“a”需要在返回表中排在第一位。 我怎樣才能做到這一點？

例如，這是雙向的，但一半的行在第 1 列和第 3 列中的順序是錯誤的。

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

我只能通過兩次調用和切換順序或 RETURN 列來獲得我想要的東西。

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

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

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)