deserialize json array list in c#

Question

I'm working on a project which has as backend mainly C#, but I'm not an experienced C# dev so I'm not able to figure out hot to fix a json deserialization of an list of objects. The following function is what takes care of the deserialization, but I get an error :

using System.IO;
using System.Web;
using Raven.Imports.Newtonsoft.Json;

namespace Corina.Web.Handlers
{
    public class JsonRequestHandler
    {
        public T Handle<T>(HttpContextBase context)
        {
            string requestData;

            context.Request.InputStream.Position = 0;
            using (var inputStream = new StreamReader(context.Request.InputStream))
            {
                requestData = inputStream.ReadToEnd();
            }

            return JsonConvert.DeserializeObject<T>(requestData, new Raven.Imports.Newtonsoft.Json.Converters.StringEnumConverter());           
        }
    }
}

Error :

Cannot deserialize the current JSON array (e.g. [1,2,3]) into type 'Corina.Web.Views.DocumentViewModel' because the type requires a JSON object (e.g. {"name":"value"}) to deserialize correctly.

To fix this error either change the JSON to a JSON object (e.g. {"name":"value"}) or change the deserialized type to an array or a type that implements a collection interface (e.g. ICollection, IList) like List that can be deserialized from a JSON array. JsonArrayAttribute can also be added to the type to force it to deserialize from a JSON array.

Can anyone tell me how do I make the deserialization on a list of objects instead of an object ?

For me what I understand is that you send together 2 JSON objects (like you have append the one to the other) and can not deserialize them together and ask you ether send one object, ether convert it to List<> — Aristos
– Aristos, Commented Feb 27, 2013 at 9:19
Well, yeah, I send something like : [ Object, Object, .... ] ; so that cannot be converted. — Roland
– Roland, Commented Feb 27, 2013 at 9:22

CodeGuru · Accepted Answer · 2013-02-27 12:48:51Z

4

You have to create this class and create a method like below :

   public class Demo
   {
      public string Name;
      public string Type;
      public string Value;
      public string ChildContentType;
      public string ChildMetadata;
   }

    public void Deserialize()
    {
        string jsonObjString = "[{\"Name\": \"Description\",\"Type\": \"Text\",\"Value\": \"XXX\",\"ChildContentType\": \"Value\",\"C??hildMetadata\": \"YYY\"}]";
         var ser = new JavaScriptSerializer();
         var arreyDemoObj = ser.Deserialize<Demo[]>(jsonObjString);

         foreach (Demo objDemo in arreyDemoObj)
         {
             //Do what you want with objDemo
         }
      }

Note that you need to add reference for JavaScriptSerializer.

answered Feb 27, 2013 at 12:48

CodeGuru

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Comments

almighty · Accepted Answer · 2013-02-27 12:26:40Z

0

Don't know the structure of your json data but i guess you are using some custom classes to deserialize with DataContractJsonSerializer you can deserialize in the following manner

Json list:

var testdata = "[{\"name\":\"numToRetrieve\",\"value\":\"3\",\"label\":\"Number of     items       to retrieve:\"},{\"name\":\"showFoo\",\"value\":\"on\",\"label\":\"Show foo?\"},  {\"name\":\"title\",\"value\":\"Foo\",\"label\":\"Foo:\"}]";

DataContractJsonSerializer js = 
new DataContractJsonSerializer(typeof (List<FooDef>));
var stream = new MemoryStream(Encoding.UTF8.GetBytes(testdata));
var foo = js.ReadObject(stream);
stream.Close();



[DataContract]
public sealed class FooDef
{
    [DataMember(Name = "name", IsRequired = true)]
    public string Name { get; set; }

    [DataMember(Name = "value", IsRequired = true)]
    public string Value { get; set; }

    [DataMember(Name = "label", IsRequired = true)]
    public string Label { get; set; }
}

answered Feb 27, 2013 at 12:26

almighty

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1 Comment

Matt Over a year ago

using System.Runtime.Serialization.Json; required for DataContractJsonSerializer. Compared to using System.Web.Script.Serialization; required for JavaScriptSerializer.

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