Question

在this处遵循本教程时，所有Gremlin Query示例都会返回动态结果。

像：

IDocumentQuery<dynamic> query = client.CreateGremlinQuery<dynamic>(graph, "g.V().hasLabel('person')");
dynamic result in await query.ExecuteNextAsync();

结果如下：

{"id":"thomas","label":"person","type":"vertex","properties":{"firstName":[{"id":"8fec3d0f-7290-4e80-9f38-9306de998579","value":"Thomas"}],"age":[{"id":"932879da-55c0-4dbc-b68c-fb3ef1e1827a","value":44}]}}

但是如何使用CreateGremlinQuery和ExecuteNextAsync来返回普通班级Person？

public class Person
{
    public string Id { get; set; }

    public string FirstName { get; set; }

    public string LastName { get; set; }

    public int Age { get; set; }
}

Answer 1

我的解决方案就像：

1] 注释人员类：

public class Person
{
    [JsonProperty("id")]
    public string Id { get; set; }

    [JsonProperty("firstName")]
    public string FirstName { get; set; }

    [JsonProperty("lastName")]
    public string LastName { get; set; }

    [JsonProperty("age")]
    public int Age { get; set; }
}

2] 将结果转换为JObject并循环属性：

IDocumentQuery<dynamic> query = client.CreateGremlinQuery<dynamic>(graph, "g.V().hasLabel('person')");
while (query.HasMoreResults)
{
    foreach (JObject item in await query.ExecuteNextAsync<dynamic>())
    {
        var properties = (JObject)item["properties"];

        var personAsJObject = new JObject();
        foreach (var property in properties)
        {
            if (property.Value is JArray valueArray)
            {
                personAsJObject.Add(property.Key, valueArray.First["value"]);
            }
        }

        var person = personAsJObject.ToObject<Person>(); // Real person here
    }
}

Microsoft.Azure.Graphs：如何将结果转换为类型？

1 个答案: