從neo4j獲取最短路徑

Question

我的任務是制定一個找到2個城鎮之間最短路程的程序。 我必須使用C＃和neo4j做到這一點。 問題是我嘗試過兩次查詢時返回的neo4j查詢的類型，問題是相同的，我找不到我必須使用哪種數據結構來接收對象。

Neo4j查詢：

MATCH  p=(a:City{name:"Pleven"})-[*]->(b:City{name:"Mezdra"})
RETURN p AS shortestPath, 
       reduce(km=0, r in relationships(p) | km+r.km) AS totalDistance
       ORDER BY totalDistance ASC
       LIMIT 1 

C＃查詢1：

CypherQuery query1 = new CypherQuery("MATCH  p=(a:City{name:'"+from+"'})-[*]->(b:City{name:'"+to+"'}) RETURN p AS shortestPath, reduce(km = 0, r in relationships(p) | km + r.km) AS totalDistance ORDER BY totalDistance ASC LIMIT 1", new Dictionary<string,object>(), CypherResultMode.Set);
            var paths = ((IRawGraphClient)client).ExecuteGetCypherResults<List<string>>(query1);

C＃查詢2：

var query1 = client.Cypher
                .Match("p=(a:City{name:{from}})-[*]->(b:City{name:{to}})")
                .WithParam("from",from)
                .WithParam("to",to)
                .Return((p) => new PathsResult<City>{
                    nodes = Return.As<IEnumerable<Node<City>>>("p AS shortestPath,reduce(km = 0, r in relationships(p) | km + r.km)"),})
                .Limit(1);
var result = query1.Results;

C＃接收包：

{"columns"frown emoticon"shortestPath","nodes"],"data":[[{"directions"frown emoticon"->","->","->"],"start":"http://localhost:7474/db/data}

Answer 1

我猜您可以只使用Cypher中的“ shortestPath”內置函數。

http://neo4j.com/docs/stable/cypherdoc-finding-paths.html

Answer 2

好的，我想出一種解決方案，我找到了一種返回城鎮之間距離的方法，現在的問題是，當我編寫p = Return.As（“ shortestPath”）時，它已被翻譯為RETURN shortestPath AS p。 當前版本的代碼：

 var query = client.Cypher
                       .Match("p=(a:City{name:{from}})-[*]->(b:City{name:{to}})")
                       .WithParam("from", from)
                       .WithParam("to", to)
                       .Return((p,order) => new
                       {
                           shortestPath = Return.As<City>("p"),
                           order= Return.As<int>("reduce(km = 0, r in relationships(p) | km + r.km)")


                       }).OrderBy("order ASC")
                       .Limit(1);

如果有任何建議請幫助

Answer 3

如果有人需要，這是完整的解決方案：

 var query = client.Cypher
                       .Match("p=(a:City{name:{from}})-[*]->(b:City{name:{to}})")
                       .WithParam("from", from)
                       .WithParam("to", to)
                       .Return((p,order) => new
                       {
                           shortestPath = Return.As<IEnumerable<Node<City>>>("nodes(p)"),
                           order= Return.As<int>("reduce(km = 0, r in relationships(p) | km + r.km)")


                       }).OrderBy("order ASC")
                       .Limit(1);

            var result = query.Results;

            foreach (var res in result) {
                foreach (var city in res.shortestPath.ToList()) {
                    Console.WriteLine(city.Data.name);
                };

            }