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Home/ Questions/Q 8608531
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Editorial Team
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Editorial Team
Asked: June 12, 20262026-06-12T03:38:09+00:00 2026-06-12T03:38:09+00:00

I know that this question has been asked ad nauseam, but the existing answers

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I know that this question has been asked ad nauseam, but the existing answers haven’t been particularly helpful for me. The best is Deserialize JSON into C# dynamic object?, but I don’t want to create an object, I want to create a Dictionary.

I serialize my data like so, and I want to deserialize to precisely the same data structure:

    public static string AugDictToJSON(Dictionary<string, List<Dictionary<string, object>>> dict)
    {
        return string.Join(",", dict.Select(
            d => string.Format("{{ \"{0}\" : [ {1} ] }}", d.Key,
                string.Join(",", d.Value.Select(i => SubAugDictToJSON(i)).ToArray())
            )).ToArray());
    }

    public static string SubAugDictToJSON(Dictionary<string, object> dict)
    {
        return string.Join(",", dict.Select(
                d => string.Format("{{ \"{0}\" : \"{1}\" }}", d.Key, d.Value.ToString())
            ).ToArray());
    }

EDIT: Solution is the following (from accepted answer):

public static Dictionary<string, List<Dictionary<string, object>>> JSONToAugDict(string json)
{
    return JsonConvert.DeserializeObject<Dictionary<string, List<Dictionary<string, object>>>>(json);
}
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  1. Editorial Team
    Editorial Team
    2026-06-12T03:38:11+00:00Added an answer on June 12, 2026 at 3:38 am

    I recommend using the NuGet package ‘Newtonsoft.Json‘.

    This package is used by WebAPI for serialization / deserialization and is quite fast. More information can be found here: http://james.newtonking.com/pages/json-net.aspx

    var results = JsonConvert.DeserializeObject<dynamic>(json);
    

    Alternatively, as an IDictionary:

    var result = JsonConvert.DeserializeObject<IDictionary<string, object>>(json);
    
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