C# json object for dynamic properties

Question

I need to output this json:

{
      white: [0, 60],
      green: [60, 1800],
      yellow: [1800, 3000],
      red: [3000, 0]
}

And I was trying to think on a model like:

 public class Colors
    {

        public int[] white { get; set; }

        public int[] green { get; set; }

        public int[] yellow { get; set; }

        public int[] red { get; set; }
    }

But the property names could change, like maybe white can be now gray, etc.

Any clue?

What about a `Dictionary`? You wouldnt need that Colors class at all — maccettura, Feb 22 '18 at 15:01
Look into DataContract classes and the System.ComponentModel namespace — Ryan Wilson, Feb 22 '18 at 15:01
You desired JSON format is invalid, you would have `"` wrapping your field names in JSON — Markiian Benovskyi, Feb 22 '18 at 15:02
you can try and read the Json and create a dynamic class based on it with `Reflection` — styx, Feb 22 '18 at 15:02
First ensure you're really producing json since that's not legal json. Your properties are missing quotes, if it is supposed to be json it has to be `"white": [0, 60], "green": [60, 1800], ` etc. Other than that, you're looking for a simple dictionary like @maccettura commented. — Lasse V. Karlsen, Feb 22 '18 at 15:05
If the names change how are you going to *read* the values? In any case, you can simply parse the string using JSON.NET and get a JObject that exposes the fields as a dictionary — Panagiotis Kanavos, Feb 22 '18 at 15:05
@styx why? JSON.Net can already return weakly-typed objects or deserialize to `dynamic` — Panagiotis Kanavos, Feb 22 '18 at 15:06
https://stackoverflow.com/questions/8368885/dynamically-adding-properties-to-a-dynamic-object — Bharadwaj, Feb 22 '18 at 15:13
@VAAA a simple JsonConvert.SerializeObject will serialize any object to a json string. — Panagiotis Kanavos, Feb 22 '18 at 15:21

maccettura · Accepted Answer · 2018-02-22T15:20:11.500

7

All you need is a Dictionary:

Dictionary<string, int[]> dictionary = new Dictionary<string, int[]>();

dictionary.Add("white", new int[] { 0, 60 });
dictionary.Add("green", new int[] { 60, 1800 });
dictionary.Add("yellow", new int[] { 1800, 3000 });
dictionary.Add("red", new int[] { 3000, 0 });

//JSON.NET to serialize
string outputJson = JsonConvert.SerializeObject(dictionary)

Results in this json:

{
    "white": [0, 60],
    "green": [60, 1800],
    "yellow": [1800, 3000],
    "red": [3000, 0]
}

Fiddle here

edited Feb 22 '18 at 15:20

answered Feb 22 '18 at 15:05

maccettura

10,514
3
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35

No need for that since Json.NET returns a weakly-typed object when parsing arbitrary strings. It's *deserialization* that requires a strongly typed object. – Panagiotis Kanavos Feb 22 '18 at 15:07
@PanagiotisKanavos I don't understand what you are saying. If JSON.NET is parsing strings, wouldnt that be _deserialization_? In which case, by your own comment, it would require a strongly typed object? OP wants to generate a specific JSON that is very clearly a string/int[] Dictionary, what solution would be better than that? – maccettura Feb 22 '18 at 15:09
A simple `JObject.ToString()` – Panagiotis Kanavos Feb 22 '18 at 15:17
@maccettura I think it would be better if you include `JsonConvert.SerializeObject(dictionary)` into the answer. Because now it only shows how to add items to dictionary and finite result – Aleks Andreev Feb 22 '18 at 15:18
1

The fiddle includes that, I can add it to the answer body. OP had a high enough reputation where I assumed they did not need help with the actual serialization, but how to make a "dynamic" object _to_ serialize – maccettura Feb 22 '18 at 15:19
@maccettura thanks for edit, upvoted. Please do not forget that your answer could be usefull not only for OP. Also link to fiddle can become broken after some time, but after your last edit your answer is self contained. – Aleks Andreev Feb 22 '18 at 15:28

score 2 · Answer 2 · 2018-02-22T17:20:02.333

2

If you don't mind using an extra library, try Json.Net (ASP.net has this pre-installed). All you have to do is

dynamic result = JsonConvert.DeserializeObject(json);

If I remember correctly, to access a value, use result[0].Value;

edited Feb 22 '18 at 17:20

answered Feb 22 '18 at 15:14

Json.NET isn't an extra library, it's the built-in default for ASP.NET Web API and the Core templates – Panagiotis Kanavos Feb 22 '18 at 15:16
@TheMCProgrammer I was partially wrong, it is built in to .NET Core, but on the other project types it is just an optional nuget package – maccettura Feb 22 '18 at 15:26
Gotcha, that makes sense. – Feb 22 '18 at 15:35

score 1 · Answer 3 · answered Feb 22 '18 at 15:23

Json.NET is the library used by almost all ASP.NET projects, including ASP.NET Web API and all ASP.NET Core projects. It can deserialize JSON to a strongly typed object or parse it to a weakly typed JObject, or generate JSON from any object. There is no need to create special classes or objects.

You can serialize any object to a Json string with JsonConvert.SerializeObject

var json=JsonConvert.SerializeObject(someObject);

Or you can use JObject as a dynamic object and convert it directly to a string :

dynamic product = new JObject();
product.ProductName = "Elbow Grease";
product.Enabled = true;
product.Price = 4.90m;
product.StockCount = 9000;
product.StockValue = 44100;
product.Tags = new JArray("Real", "OnSale");

Console.WriteLine(product.ToString());

C# json object for dynamic properties

3 Answers3

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