用于将节点从一个列表结构移动到另一个列表结构的Cypher查询
用于将节点从一个列表结构移动到另一个列表结构的Cypher查询
提问于 2020-03-11 16:39:34
我们有以下结构，其中用户是起点，带有数字的节点被标记为邀请，其值指定它们的属性id。
我正在寻找一种创建查询的方法，该查询将节点从VALID_INVITATIONS关系指向的列表移动到由INVALID_INVITATIONS指向的另一个列表。移动节点应首先在新列表中设置。
我想出了一个可行的解决方案，但由于缺乏对Cypher的知识和经验，我恳请社区为之提供帮助和改进。正如您将看到的那样，有许多命令式代码“hack”(情况下)，而不是像Cypher所认为的那样是声明性的。我很感谢你的提示和建议。
MATCH (u:User)
MATCH (u)-[:VALID_INVITATIONS|PREVIOUS*]->(current:Invitation{ id:3 })
//find (current) node's predecessor which might pointing on it through VALID_INVITATIONS or PREVIOUS relationships
OPTIONAL MATCH (u)-[oldValidRel:VALID_INVITATIONS]->(current)
OPTIONAL MATCH (predecessor:Invitation)-[oldPredecessorRel:PREVIOUS]->(current)
//find (current) node's subsequent node
OPTIONAL MATCH (current)-[oldSubsequentRel:PREVIOUS]->(subsequent:Invitation)
//first we re-create connections in list pointed by VALID_INVITATION relationship for consistency
WITH *, CASE subsequent WHEN NULL THEN [] ELSE [1] END AS hasSubsequent
//if (current) node is connected to (u) User we replace VALID_INVITATIONS relationship
FOREACH(_ IN CASE oldValidRel WHEN NULL THEN [] ELSE [1] END |
DELETE oldValidRel
//if (subsequent) node exist then we need to re-link it
FOREACH(_ IN hasSubsequent |
MERGE (u)-[:VALID_INVITATIONS]->(subsequent)
DELETE oldSubsequentRel
)
)
//following condition should be XOR with the one above (only one must be executed)
//if (current) node has another Invitation node as predecessor in list
FOREACH(_ IN CASE oldPredecessorRel WHEN NULL THEN [] ELSE [1] END |
DELETE oldPredecessorRel
//if (subsequent) node exist then we need to re-link it
FOREACH(_ IN hasSubsequent |
MERGE (predecessor)-[:PREVIOUS]->(subsequent)
DELETE oldSubsequentRel
)
)
WITH u, current
//now it is time to move (current) node to beginning of the list pointed by INVALID_INVITATIONS relationship
MERGE (u)-[:INVALID_INVITATIONS]->(current)
WITH u, current
//find obsolete oldRel:INVALID_INVITATIONS relationship
MATCH (current)<-[:INVALID_INVITATIONS]-(u)-[oldRel:INVALID_INVITATIONS]->(oldInv)
DELETE oldRel
//link (current) with previously "first" node in INVALID_INVITATIONS list
MERGE (current)-[:PREVIOUS]->(oldInv)回答 1
Stack Overflow用户
回答已采纳
发布于 2020-03-13 08:00:16
在享受了一些“乐趣”和assisstance from @cybersam之后，我想出了一个更短的解决方案：
MATCH (u:User)-[:VALID_INVITATIONS|PREVIOUS*]->(current:Invitation {id:4})
MATCH (predecessor)-[oldPredecessorRel:VALID_INVITATIONS|PREVIOUS]->(current)
OPTIONAL MATCH (current)-[oldSubsequentRel:PREVIOUS]->(subsequent)
CALL apoc.do.when(subsequent IS NOT NULL, "CALL apoc.create.relationship($p, $r, NULL, $s) YIELD rel RETURN rel", "", {p:predecessor, r:type(oldPredecessorRel), s:subsequent}) YIELD value
DELETE oldPredecessorRel, oldSubsequentRel
WITH u, current
MERGE (u)-[:INVALID_INVITATIONS]->(current)
WITH u, current
MATCH (oldInv)<-[oldRel:INVALID_INVITATIONS]-(u)-[:INVALID_INVITATIONS]->(current)
DELETE oldRel
MERGE (current)-[:PREVIOUS]->(oldInv)页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接：
https://stackoverflow.com/questions/60640755
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