c# - 将最短路径中的所有节点作为对象列表返回

我有以下Cypher Query，在Neo4j 2.0.0中可以正常工作。

MATCH (ab:Point { Latitude: 24.96325, Longitude: 67.11343 }),(cd:Point { Latitude: 24.95873, Longitude: 67.10335 }),
p = shortestPath((ab)-[*..150]-(cd))
RETURN p

Neo4jClient中的以下查询给出了错误:函数评估超时。

var pathsQuery =
            client.Cypher
            .Match("(ab:Point { Latitude: 24.96325, Longitude: 67.11343 }),(cd:Point { Latitude: 24.95873, Longitude: 67.10335 }), p = shortestPath((ab)-[*..150]-(cd))")
            .Return<IEnumerable<PointEntity>>("extract(n in nodes(p) : id(n))");

并且在Xclave上进行了另一项类似的工作之后:

http://geekswithblogs.net/cskardon/archive/2013/07/23/neo4jclient-ndash-getting-path-results.aspx

结果值为:功能评估超时。

var pathsQuery =
            client.Cypher
            .Match("(ab:Point { Latitude: 24.96325, Longitude: 67.11343 }),(cd:Point { Latitude: 24.95873, Longitude: 67.10335 }), p = shortestPath((ab)-[*..150]-(cd))")
            .Return(p => new PathsResult<PointEntity>
                {
                    Nodes = Return.As<IEnumerable<Node<PointEntity>>>("nodes(p)"),
                });

如何获取C#中查询返回的所有节点作为PointEntity Types列表？

编辑:

我能够从以下代码中获取节点和关系的URI:

var pathsQuery =
            client.Cypher
            .Match("(ab:Point { Latitude: 24.96325, Longitude: 67.11343 }),(cd:Point { Latitude: 24.95873, Longitude: 67.10335 }), p = shortestPath((ab)-[*..150]-(cd))")
var results = pathsQuery.Return<PathsResult>("p").Results;

在以下Craig Brett的博客之后:
http://craigbrettdevden.blogspot.co.uk/2013/03/retrieving-paths-in-neo4jclient.html

我试图获取POCO，但出现错误:

var paths = pathsQuery.Returns<PathsResult>("EXTRACT(n in nodes(p) : n) AS Nodes, EXTRACT(rel in rels(p) : rel) AS Relationships", CypherResultMode.Projection).Results;

错误:

'Neo4jClient.Cypher.ICypherFluentQuery' does not contain a definition for 'Returns' and no extension method 'Returns' accepting a first argument of type 'Neo4jClient.Cypher.ICypherFluentQuery' could be found (are you missing a using directive or an assembly reference?)

最佳答案

我不能说为什么第一个查询不起作用，我现在无法对其进行测试，因为我的Neo4j版本不允许我在MATCH子句中使用参数。但是-我确实得到了我的版本(如下)响应良好。在超时方面，[*..150]是一个很大的潜在遍历对象。

你可以尝试运行:

var pathsQuery =
    Client.Cypher
        .Match("p = shortestPath((ab:Point)-[*..150]-(cd:Point))")
        .Where((PointEntity ab) => ab.Latitude == 24.96325)
        .AndWhere((PointEntity ab) => ab.Longitude == 67.11343)
        .AndWhere((PointEntity cd) => cd.Latitude == 24.95873)
        .AndWhere((PointEntity cd) => cd.Longitude == 67.10335)

        .Return(p => new PathsResult<PointEntity>
                        {
                            Nodes = Return.As<IEnumerable<Node<PointEntity>>>("nodes(p)"),
                            Relationships = Return.As<IEnumerable<RelationshipInstance<object>>>("rels(p)")
                        });

var res = pathsQuery.Results;

其中PathsResult定义为:

public class PathsResult<TNode>
{
    public IEnumerable<Node<TNode>> Nodes { get; set; }
    public IEnumerable<RelationshipInstance<object>> Relationships { get; set; }
}

这样做的好处是WHERE子句是参数化的。