Question

我有以下Cypher查询，它在Neo4j 2.0.0中运行良好。

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中的以下查询给出错误：功能评估已超时。

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))");

AND跟随另一个关于Xclave的类似工作：

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)"),
                });

如何将C＃中查询返回的所有节点作为PointEntity类型列表获取？

编辑：

我能够从这段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;

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

我试图获得POCO但却出错：

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?)

Answer 1

我不能说为什么第一个查询不起作用，我现在无法测试它，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（下面）很好。就超时而言，[*..150]是一个非常大的潜在遍历。

你可以尝试跑步吗？

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;

PathsResult定义为：

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

这里的好处是WHERE条款是绝对的。

将最短路径中的所有节点作为对象列表返回

1 个答案: