从查询 neo4j 的结果创建新关系

Question

我想从一个节点和一组节点之间的最短路径的结果中建立新的关系。我目前正在使用此查询来执行此类任务，但在两个节点之间（效果很好）：

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)

我如何概括此查询以返回 p2 ("Monarchs_of_the_Bulgars") 和节点列表 p1 = ["Health", "History", "Humanity"] 之间的 allshortestPaths 列表，而不是像上一个查询？

感谢您的帮助。

Answer 1

这应该可行：

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)

[更新]

如果p1 也必须是Top_Main_Classification 的直接子目录，那么您可以这样做：

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)