今天看到社区有人提问如何进行关系重构，顺手回答了一下。在此记录下关系重构的方法。
- 创建测试数据
代码语言：javascript复制MERGE (A:Test {name:'A'})
MERGE (B:Test {name:'B'})
CREATE (A)-[:Realation {name:'属性1'}]->(B)
CREATE (A)-[:Realation {name:'属性1'}]->(B)
CREATE (A)-[:Realation {name:'属性2'}]->(B)
CREATE (A)-[:Realation {name:'属性2'}]->(B)
CREATE (A)-[:Realation {name:'属性2'}]->(B)
CREATE (A)-[:Realation {name:'属性3'}]->(B)- 查询测试数据
代码语言：javascript复制MATCH p=(A:Test {name:'A'})-->(B:Test {name:'B'}) RETURN p
- 如何重构
代码语言：javascript复制想请教下大佬，如何删除两个节点间的重复关系，重复的定义指的是，关系的属性不同
比如
(A)-[:Realation{name:‘属性1’]-(B)
(A)-[:Realation{name:‘属性1’]-(B)
(A)-[:Realation{name:‘属性2’]-(B)
(A)-[:Realation{name:‘属性2’]-(B)
(A)-[:Realation{name:‘属性2’]-(B)
(A)-[:Realation{name:‘属性3’]-(B)
想把重复的部分去掉，就是变成
(A)-[:Realation{name:‘属性1’]-(B)
(A)-[:Realation{name:‘属性2’]-(B)
(A)-[:Realation{name:‘属性3’]-(B)- 重构关系
代码语言：javascript复制MATCH p=(A:Test {name:'A'})-[r]->(B:Test {name:'B'})
WITH ID(r) AS id,r.name AS name
WITH name,COLLECT(id) AS relIds
WITH name,relIds,SIZE(relIds) AS relIdsSize
WHERE relIdsSize>1
WITH name,apoc.coll.subtract(relIds, [relIds[0]]) AS deleteRelIds
WITH name,deleteRelIds
MATCH ()-[r]-() WHERE ID(r) IN deleteRelIds DELETE r- 重构结果
- 更多复杂重构可以使用下面的存储过程实现
代码语言：javascript复制CALL apoc.do.case([relationship=1,'MATCH (from:Label {hcode:$fromHcode}),(to:Label {hcode:$toHcode}) MERGE (from)-[:NEXT]->(to)',relationship=-1,'MATCH (from:Label {hcode:$fromHcode}),(to:Label {hcode:$toHcode}) MERGE (from)<-[:NEXT]-(to)'],'',{fromHcode:fromHcode,toHcode:toHcode}) YIELD value RETURN value- 社区问答连接
代码语言：javascript复制http://neo4j.com.cn/topic/5f3b28e4a4477ec754d2b55f- 推荐一下lyonwj的博客
代码语言：javascript复制https://www.lyonwj.com/
