如何使用newtonsoft json解析3级嵌套数组

时间:2015-07-10 12:22:03

标签: c# json json.net

我有以下json,它采用geojson格式,并且Id希望能够将其解析为c#中的嵌套列表:

public IList<IList<IList<double>>> Coordinates { get; set; }

"coordinates": [
    [
        [-3.213338431720785, 55.940382588499197],
        [-3.213340490487523, 55.940381867350276],
        [-3.213340490487523, 55.940381867350276],
        [-3.213814166228732, 55.940215021175085],
        [-3.21413960035129, 55.940100842843712]
    ]
]

我尝试了以下但是我得到了一个例外:

    var node = jsonProperties["geometry"]["coordinates"].Values();
    var coordinates = node.Select(x=>x.Value<List<double>>());

例外细节:

  

无法将Newtonsoft.Json.Linq.JArray转换为   Newtonsoft.Json.Linq.JToken。

3 个答案:

答案 0 :(得分:2)

使用newtonsoft反序列化。使用Foo属性创建coordinates类,使用大括号括起JSON脚本以将其表示为对象,然后调用JsonConvert.DeserializeObject<Foo>(Json)

private class Foo
{
     public List<List<List<double>>> coordinates { get; set; }
}

 var json = @"{
                 coordinates: [
                                [
                                    [-3.213338431720785, 55.940382588499197],
                                    [-3.213340490487523, 55.940381867350276],
                                    [-3.213340490487523, 55.940381867350276],
                                    [-3.213814166228732, 55.940215021175085],
                                    [-3.21413960035129, 55.940100842843712]
                                ]
                            ]
                    }";
var result = JsonConvert.DeserializeObject<Foo>(json);

答案 1 :(得分:1)

可能不是您想要的,但使用动态类型我可以访问这些值。例如,此示例代码

class Program
{
    static void Main(string[] args)
    {
        string sampleJson = @"{ ""coordinates"": [
            [
                [-3.213338431720785, 55.940382588499197],
                [-3.213340490487523, 55.940381867350276],
                [-3.213340490487523, 55.940381867350276],
                [-3.213814166228732, 55.940215021175085],
                [-3.21413960035129, 55.940100842843712]
            ]
        ]}";

        dynamic d = JObject.Parse(sampleJson);

        Console.WriteLine(d.coordinates[0].Count);
        foreach (var coord in d.coordinates[0])
        {
            Console.WriteLine("{0}, {1}", coord[0], coord[1]);
        }

   Console.ReadLine();
}

显示以下内容:

5
-3.21333843172079, 55.9403825884992
-3.21334049048752, 55.9403818673503
-3.21334049048752, 55.9403818673503
-3.21381416622873, 55.9402150211751
-3.21413960035129, 55.9401008428437

答案 2 :(得分:1)

我建议您将它们解析为更合适的内容,例如List<Tuple<double, double>>,尽管还有嵌套列表的解决方案。请查看我的内联评论:

const string json = @"
    { 
        ""coordinates"": 
        [
            [
                [-3.213338431720785, 55.940382588499197],
                [-3.213340490487523, 55.940381867350276],
                [-3.213340490487523, 55.940381867350276],
                [-3.213814166228732, 55.940215021175085],
                [-3.21413960035129, 55.940100842843712]
            ]
        ]
    }";

var jsObject = JObject.Parse(json);
/*
 * 1. Read property "coordinates" of your root object
 * 2. Take first element of array under "coordinates"
 * 3. Select each pair-array and parse their values as doubles
 * 4. Make a list of it
 */
var result = jsObject["coordinates"]
                .First()
                .Select(pair => new Tuple<double, double> (
                        pair[0].Value<double>(), 
                        pair[1].Value<double>()
                    )
                ).ToList();

对于List<List<List<double>>>,请参阅@YTAM回答。