Neo4j最短路徑問題
1.指定某一結點
- 無向邊：
MATCH (p1:Person {name:"aaaaaaa"}),(p2:Person{name:"bbbbbb"}),
p=shortestpath((p1)-[*..10]-(p2))
RETURN p
- 有向邊：
MATCH (p1:Person {name:"aaaaaaa"}),(p2:Person{name:"bbbbbb"}),
p=shortestpath((p1)-[*..10]->(p2))
RETURN p
注：[*…10]表示查詢路徑長度10以內的關系
- 同時返回最短路徑長度：
MATCH (p1:Person {name:"aaaaaaa"}),(p2:Person{name:"bbbbbb"}),
p=shortestpath((p1)-[*..10]->(p2))
RETURN p,length(p)
- 添加限制條件
- 1）只經過標簽為“rrrr”的邊:
MATCH (p1:Person {name:"aaaaaaa"}),(p2:Person{name:"bbbbbb"}),
p=shortestpath((p1)-[r:rrrr*..10]->(p2))
RETURN p
- 2）不經過屬性值idp為"xxxx"的結點:
MATCH (p1:Person {name:"aaaaaaa"}),(p2:Person{name:"bbbbbb"}),
p=shortestpath((p1)-[*..10]->(p2))
where all(x in nodes(p) where x.idp<>"xxxx")
RETURN p
- 2）不經過屬性值idr為"yyyy"的邊:
MATCH (p1:Person {name:"aaaaaaa"}),(p2:Person{name:"bbbbbb"}),
p=shortestpath((p1)-[r*..10]->(p2))
where all(x in r where x.idr<>"yyyy")
RETURN p
2.指定某一類結點
(邊的有向無向、限制條件同上，此處不再分別敘述)
match (p:Person) with collect(p) as nodes
unwind nodes as source
unwind nodes as target
with source,target where id(source)<>id(target)
match paths = shortestPath((source)-[*..10]->(target))
with paths limit 25
return path
- 返回所有最短路徑
match (p:Person) with collect(p) as nodes
unwind nodes as source
unwind nodes as target
with source,target where id(source)<>id(target)
match paths = allShortestPaths((source)-[*..10]->(target))
with paths limit 25
return path
with source,target where id(source)<>id(target)
此處是為了保證起始和最終結點不相同。
注：兩個unwind把結點集合打散，並以笛卡爾積的形式組成結點對。
參考：https://blog.csdn.net/wry2008wry/article/details/80762811
原文地址：https://blog.csdn.net/qq_34233510/article/details/83110854 </div>