Question

我有一个任务是制作一个在两个城镇之间找到最短路的程序。我不得不通过使用C＃和neo4j来实现它。问题是返回类型的neo4j查询我尝试了两个查询，问题是相同的，我无法找到我必须使用什么样的数据结构来接收对象。

Neo4j查询：

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＃查询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＃查询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＃收到包

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

Answer 1

我猜你可以在Cypher中使用“shortestPath”内置函数。

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

Answer 2

好的，我找到了一种解决方案，我找到了一种方法来恢复城镇之间的距离，现在的问题是，当我写下p = Return.As（＆＃34; shortestPath＆＃34;）它已被翻译为RETURN shortestPath AS p。当前版本的代码：

 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);

如有任何建议请帮助

Answer 3

如果有人需要，那就是完整的解决方案：

 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);
                };

            }

从neo4j获取最短路径

3 个答案: