Question

我想根据一个节点和一组节点之间最短路径的结果建立新的关系。我目前正在使用此查询来执行此类任务，但要在两个节点之间进行（很好）：

MATCH
  (p1:Category {catName: "Main_topic_classifications"}), 
  (p2:Category {catName: "Monarchs_of_the_Bulgars"}),
  path = allShortestPaths((p2)-[:SUBCAT_OF*]->(p1))
UNWIND RELATIONSHIPS(path) AS rel
WITH STARTNODE(rel) AS s, ENDNODE(rel) AS e
MERGE (s)-[:NEW_SUBCAT]->(e)

如何归纳此查询以返回p2（“ Monarchs_of_the_Bulgars”）和节点列表p1 = [“ Health”，“ History”，“ Humanity”]之间的所有最短路径列表，而不是返回像上一个查询？

感谢您的帮助。

Answer 1

这应该有效：

MATCH (p1:Category)
WHERE p1.category IN ["Health", "History", "Humanity"]
MATCH
  (p2:Category {catName: "Monarchs_of_the_Bulgars"}),
  path = allShortestPaths((p2)-[:SUBCAT_OF*]->(p1))
UNWIND RELATIONSHIPS(path) AS rel
WITH STARTNODE(rel) AS s, ENDNODE(rel) AS e
MERGE (s)-[:NEW_SUBCAT]->(e)

[已更新]

如果p1也必须是Top_Main_Classification的直接子目录，则可以执行以下操作：

MATCH (p1:Category)-[:SUBCAT_OF]->(main:Category {catName: "Main_topic_classifications"})
WHERE p1.category IN ["Health", "History", "Humanity"]
MERGE (p1)-[:NEW_SUBCAT]->(main)
MATCH
  (p2:Category {catName: "Monarchs_of_the_Bulgars"}),
  path = allShortestPaths((p2)-[:SUBCAT_OF*]->(p1))
UNWIND RELATIONSHIPS(path) AS rel
WITH STARTNODE(rel) AS s, ENDNODE(rel) AS e
MERGE (s)-[:NEW_SUBCAT]->(e)

根据查询neo4j

1 个答案: