Well, if you just need a single extra property, one simple approach is to parse your JSON to a JObject, use ToObject() to populate your class from the JObject, and then use SelectToken() to pull in the extra property. So, assuming your class looked something like this: class Person { [JsonProperty(“name”)] public string Name { … Read more
For simple array members like that, you can use JSON.parse. var array = JSON.parse(“[” + string + “]”); This gives you an Array of numbers. [0, 1] If you use .split(), you’ll end up with an Array of strings. [“0”, “1”] Just be aware that JSON.parse will limit you to the supported data types. If … Read more
I’d take a slightly different approach as follows, so as to minimize “manual” parsing in my code, as unnecessarily doing otherwise somewhat defeats the purpose of why I’d use an API like Gson in the first place. // output: // [User: id=1, name=Jonas, updateDate=2011-03-24 03:35:00.226] // [User: id=5, name=Test, updateDate=2011-05-07 08:31:38.024] // using java.sql.Timestamp public … Read more
You can use the C# dynamic type to make things easier. This technique also makes re-factoring simpler as it does not rely on magic-strings. JSON The JSON string below is a simple response from an HTTP API call, and it defines two properties: Id and Name. {“Id”: 1, “Name”: “biofractal”} C# Use JsonConvert.DeserializeObject<dynamic>() to deserialize … Read more
Json.NET – Newtonsoft has a JsonPropertyAttribute which allows you to specify the name of a JSON property, so your code should be: public class TeamScore { [JsonProperty(“eighty_min_score”)] public string EightyMinScore { get; set; } [JsonProperty(“home_or_away”)] public string HomeOrAway { get; set; } [JsonProperty(“score “)] public string Score { get; set; } [JsonProperty(“team_id”)] public string TeamId … Read more
If you just want a generic method that can handle any arbitrary JSON and convert it into a nested structure of regular .NET types (primitives, Lists and Dictionaries), you can use JSON.Net’s LINQ-to-JSON API to do it: using System.Linq; using Newtonsoft.Json.Linq; public static class JsonHelper { public static object Deserialize(string json) { return ToObject(JToken.Parse(json)); } … Read more
Use var rootObject = JsonConvert.DeserializeObject<RootObject>(string json); Create your classes on JSON 2 C# Json.NET documentation: Serializing and Deserializing JSON with Json.NET
Using the standard CustomCreationConverter, I was struggling to work how to generate the correct type (Person or Employee), because in order to determine this you need to analyse the JSON and there is no built in way to do this using the Create method. I found a discussion thread pertaining to type conversion and it … Read more
As was answered here – Deserialize JSON into C# dynamic object? It’s pretty simple using Json.NET: dynamic stuff = JsonConvert.DeserializeObject(“{ ‘Name’: ‘Jon Smith’, ‘Address’: { ‘City’: ‘New York’, ‘State’: ‘NY’ }, ‘Age’: 42 }”); string name = stuff.Name; string address = stuff.Address.City; Or using Newtonsoft.Json.Linq : dynamic stuff = JObject.Parse(“{ ‘Name’: ‘Jon Smith’, ‘Address’: { … Read more