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无法将当前 JSON 对象(例如 {“name":“value"})反序列化为类型“System.Collections.Generic.List"1

更新时间:2022-10-30 11:06:16

说清楚,除了@SLaks 的回答之外,这意味着您需要更改这一行:

Listdatalist = JsonConvert.DeserializeObject>(jsonstring);

像这样:

RootObject datalist = JsonConvert.DeserializeObject(jsonstring);

using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
using Facebook;
using Newtonsoft.Json;

namespace facebook
{
    class Program
    {
        static void Main(string[] args)
        {
            var client = new FacebookClient(acc_ess);
            dynamic result = client.Get("fql", new { q = "select target_id,target_type from connection where source_id = me()"});
            string jsonstring = JsonConvert.SerializeObject(result);

            //jsonstring {"data":[{"target_id":9503123,"target_type":"user"}]}
            List<RootObject> datalist = JsonConvert.DeserializeObject<List<RootObject>>(jsonstring);
        }

        public class Datum
        {
            public Int64 target_id { get; set; }
            public string target_type { get; set; }
        }

        public class RootObject
        {          
            public List<Datum> data { get; set; }
        }
    }
}

Cannot deserialize the current JSON object (e.g. {"name":"value"}) into type 'System.Collections.Generic.List`1[facebook.Program+RootObject]' because the type requires a JSON array (e.g. [1,2,3]) to deserialize correctly. To fix this error either change the JSON to a JSON array (e.g. [1,2,3]) or change the deserialized type so that it is a normal .NET type (e.g. not a primitive type like integer, not a collection type like an array or List) that can be

I looked at other posts.

My json looks like this:

{"data":[{"target_id":9503123,"target_type":"user"}]}

To make it clear, in addition to @SLaks' answer, that meant you need to change this line :

List<RootObject> datalist = JsonConvert.DeserializeObject<List<RootObject>>(jsonstring);

to something like this :

RootObject datalist = JsonConvert.DeserializeObject<RootObject>(jsonstring);