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以下代码返回整个 JSON 对象,但我需要过滤到某些特定对象,例如我只需要获取"layers"或者tiltle请让我知道如何做到这一点是 C#?

我必须为此创建一个HttpWebRequest对象吗?如果是这样,我应该在哪里传递请求的数据?

using (WebClient wc = new WebClient())
   {
    var json = wc.DownloadString("https://sampleserver6.arcgisonline.com/arcgis/rest/services/Water_Network/FeatureServer?f=pjson");
    Console.WriteLine(json);
   }

我已经尝试过了,但它也返回了所有内容

class Program
{
    private const string URL = "https://sampleserver6.arcgisonline.com/arcgis/rest/services/Water_Network/FeatureServer?f=pjson";
    private const string DATA = @"{{""layers"":""Layers""}}";
    static void Main(string[] args)
    {

       CreateObject();

    }
    private static void CreateObject()
    {
        HttpWebRequest request = (HttpWebRequest)WebRequest.Create(URL);
        request.Method = "POST";
        request.ContentType = "application/json";
        request.ContentLength = DATA.Length;
        using (Stream webStream = request.GetRequestStream())
        using (StreamWriter requestWriter = new StreamWriter(webStream, System.Text.Encoding.ASCII))
        {
            requestWriter.Write(DATA);
        }

        try
        {
            WebResponse webResponse = request.GetResponse();
            using (Stream webStream = webResponse.GetResponseStream())
            {
                if (webStream != null)
                {
                    using (StreamReader responseReader = new StreamReader(webStream))
                    {
                        string response = responseReader.ReadToEnd();
                        Console.WriteLine(response);
                        Console.ReadLine();
                    }
                }
            }
        }
        catch (Exception e)
        {
            Console.WriteLine("-----------------");
            Console.WriteLine(e.Message);
            Console.ReadLine();
        }

    }

}
4

1 回答 1

2

如果您需要与layers数组对象相关的信息,那么您可以使用以下代码

 using (var wc = new WebClient())
 {
     string json = wc.DownloadString("https://sampleserver6.arcgisonline.com/arcgis/rest/services/Water_Network/FeatureServer?f=pjson");
     dynamic data = Json.Decode(json);
     Console.WriteLine(data.layers[0].id);
     Console.WriteLine(data.layers[0].name);
     Console.WriteLine(data.documentInfo.Title);
 }
于 2017-03-31T16:59:28.837 回答