Question

假设这种拓扑结构：

create
 (mbp {name: "MacBookPro"}),
 (fb {name:"FritzBox"}),
 (eue {name: "1u1"}),
 (inet {name:"Internet"}),
 (mnh {name:"Mannheim"}),
 (hdb {name:"Heidelberg"}),
 (stg {name: "Stuttgart"}),
 (lnd {name:"Lindt"}),
 (ylo {name:"Ylo"}),
 (qvin {name:"Qvin"}),
 (mbp) -[:LNK {bw:100}]->(fb),
 (fb)-[:LNK {bw:32}]->(eue),
 (eue)-[:LNK {bw:10000}]->(inet),
 (inet)-[:LNK {bw:10000}]->(mnh),
 (mnh)-[:LNK {bw:10000}]->(hdb),
 (hdb)-[:LNK {bw:1000}]->(lnd),   // Route option 1
 (mnh)-[:LNK {bw:10000}]->(stg),
 (stg)-[:LNK {bw:10000}]->(lnd),  // Route option 2
 (lnd)-[:LNK {bw:1000}]->(ylo),
 (lnd)-[:LNK {bw:1000}]->(qvin),
 (ylo)-[:LNK {bw:1000}]->(qvin)

我可以使用此查询获取“MacBookPro”和“Ylo”之间的最大带宽路径：

match (s {name:"MacBookPro"}),
  (t {name:"Ylo"}),
  p=allShortestPaths(s-[*]-t)
  return
    p as path,
    reduce( mini=100000,
            r in relationships(p) |
              case when mini < r.bw then mini else r.bw end)
        as maximal_bandwidth
  order by maximal_bandwidth desc
  limit 1;

如何在该路径的所有节点上创建新关系（例如SESSION_RTP）？

根据Snowburnts的建议，我想出了这个部分解决方案：

match (s {name:"MacBookPro"}),
  (t {name:"Ylo"}),
   p=allShortestPaths(s-[*]-t)
   with
   p as path,
   reduce( mini=100000,
           r in relationships(p) |
             case when mini < r.bw then mini else r.bw end)
        as maximal_bandwidth
  order by maximal_bandwidth desc limit 1
  foreach( r in relationships(path)| set r.session = 1 )

其次是：

match (a)-[r:LNK {session : 1}]->(b) create (a)-[:SESSION_RTP{id:1}]->(b)

是否可以在一个声明中压缩它？

Answer 1

阅读评论后

我知道只有bw最小的路径才能将节点与名为SESSION_RIP的关系连接起来。

尝试以下

match (s {name:"MacBookPro"}),
  (t {name:"Ylo"}),
   p=allShortestPaths(s-[*]-t)
   with
   p as path,
   reduce( mini=100000,
           r in relationships(p) |
             case when mini < r.bw then mini else r.bw end)
        as maximal_bandwidth
  order by maximal_bandwidth desc limit 1
  with relationships(p) as rels
  foreach (t in rels | 
  foreach (a in [startNode(t)]| 
  foreach (b in [endNode(t)]| 
  CREATE a-[:SESSION_RTP]->b
  )))

沿路径创建新关系

1 个答案: