返回最短路径中的所有节点作为对象列表 [英] Return All Nodes in Shortest Path as Object List

问题描述

我有以下 Cypher 查询，它在 Neo4j 2.0.0 中运行良好.

I have the following Cypher Query which works fine in the Neo4j 2.0.0.

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:Function Evaluation Timed.

The following Query in Neo4jClient gives the error: Function Evaluation Timed out.

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 在 Xclave 上进行了另一项类似的工作之后:

AND After following Another Similar work on Xclave:

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

结果值为:函数评估超时.

The Results value as : Function Evaluation Timed out.

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 类型列表?

How can I get all the nodes returned by the query in C# as a List of PointEntity Types?

我能够从这段代码中取回节点和关系的 URI:

I was able to get back the URIs of the nodes and relations from this code:

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;

在此处关注 Craig Brett 的博客:http://craigbrettdevden.blogspot.co.uk/2013/03/retrieving-path-in-neo4jclient.html

Following Craig Brett's blog here: http://craigbrettdevden.blogspot.co.uk/2013/03/retrieving-paths-in-neo4jclient.html

我尝试获取 POCO 但出现错误:

I tried to get the POCO but got error:

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?)

推荐答案

我不知道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 版本不允许我使用我的 MATCH 子句中的参数.但是 - 我的版本(如下)响应良好.在超时方面，[*..150] 是一个相当大的潜在遍历.

I can't say why the first query isn't working, I can't test it right now as my version of Neo4j isn't letting me use parameters in my MATCH clause. However - I do get my version (below) responding fine. In terms of timing out, the [*..150] is quite a large potential traversal.

你可以试试跑步吗:

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;

其中 PathsResult 定义为:

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

这里的好处是 WHERE 子句是参数化的.

The benefit here is that the WHERE clauses are paramaterized.

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