JSON change value

Question

I have a JSON value like this

{"$id":"649271776","$type":"outdoorgame","Overs":50,"Balls":6,"TeamName":"TestTeam"}

I wrote a C# code like this to change the value of Overs from 50 to 10

var jsonString = sSession.GameState; //this is the value {"$id":"649271776","$type":"outdoorgame","Overs":50,"Balls":6,"TeamName":"TestTeam"}
dynamic jsonObject = 
Newtonsoft.Json.JsonConvert.DeserializeObject(jsonString);
jsonObject.Overs = 10;
var modifiedJsonString = Newtonsoft.Json.JsonConvert.SerializeObject(jsonObject);

This code is changing the value of Overs from 50 to 10. The problem I am facing when I use the above code modifiedJsonString is missing these two values

"$id":"649271776","$type":"outdoorgame"

giving the output as {Overs":10,"Balls":6,"TeamName":"TestTeam"} I want $id and $type also in the modifiedJsonString.

I want modifiedJsonString like this {"$id":"649271776","$type":"outdoorgame","Overs":10,"Balls":6,"TeamName":"TestTeam"}

Can anyone tell me how to solve this problem

Actually JSON is correct only, I guess something has to do with $id and $type, so have to do the code differently (not sure) — Arul Joseph, May 30 '18 at 16:21
After modifying the overs I am Serialising like this var modifiedJsonString = Newtonsoft.Json.JsonConvert.SerializeObject(jsonObject); — Arul Joseph, May 30 '18 at 16:23
Before Serializing the jsonObject the values are correct, after the Serializing the jsonObject then only $id and $type disappears — Arul Joseph, May 30 '18 at 16:24
Where does this JSON come from in the first place? How is it created? — DavidG, May 30 '18 at 16:26
Check out https://stackoverflow.com/questions/24218536/deserialize-json-that-has-some-property-name-starting-with-a-number, when you deserialize the object those properties disappear because they start with `$` — Yuriy Faktorovich, May 30 '18 at 16:26
Actually JSON values come from the client application, I request the server with the above values to change the Overs, after changing server will send it back to client. In server I am storing all the actions of a game — Arul Joseph, May 30 '18 at 16:29
You miss my point, how is this JSON created by the server? Do you have that code? — DavidG, May 30 '18 at 16:42

score 2 · Answer 1 · answered May 30 '18 at 17:24

The problem is that $id and $type are not valid identifiers, and can't appear as members of the returned dynamic object built by the JSON serializer. As in gldraphael's answer, the solution is to create your own concrete class to hold the deserialized object; for the properties whose names start with $ you'll need to use JsonPropertyAttribute to remap the names:

public class GameState
{
    [JsonProperty("$id")] public string ID { get; set; }
    [JsonProperty("$type")] public string Type { get; set; }
    int Overs { get; set; }
    int Balls { get; set; }
    public string TeamName { get; set; }
}

Further, Json.NET treats $type as a special property name and this interferes with proper deserialization of your object. To get around this, we must use the MetadataPropertyHandling.Ignore serializer setting.

Thus you can deserialize, modify and re-serialize like this:

string jsonString = "{\"$id\":\"649271776\",\"$type\":\"outdoorgame\",\"Overs\":50,\"Balls\":6,\"TeamName\":\"TestTeam\"}";
JsonSerializerSettings settings = new JsonSerializerSettings() { MetadataPropertyHandling = MetadataPropertyHandling.Ignore };
GameState jsonObject = JsonConvert.DeserializeObject<GameState>(jsonString, settings);
jsonObject.Overs = 10;
var modifiedJsonString = JsonConvert.SerializeObject(jsonObject);

See it in action.

Stringfellow · Answer 2 · 2018-05-30T17:00:48.213

1

You can use JToken to handle this.

using Newtonsoft.Json; using Newtonsoft.Json.Linq;

var jsonString = "{\"$id\":\"649271776\",\"$type\":\"outdoorgame\",\"Overs\":50,\"Balls\":6,\"TeamName\":\"TestTeam\"}";
JToken jsonObject = JToken.Parse(jsonString);
jsonObject["Overs"] = 10;
var modifiedJsonString = JsonConvert.SerializeObject(jsonObject);

// In case one wanted to update the $type and $id fields
jsonObject["$type"] = "asdf";
jsonObject["$id"] = 123456;
var modifiedJsonString2 = JsonConvert.SerializeObject(jsonObject);

Will result in: modifiedJsonString --> {"$id":"649271776","$type":"outdoorgame","Overs":10,"Balls":6,"TeamName":"TestTeam"}

And if you needed to update $id and $type, that is possible, too. modifiedJsonString2 --> {"$id":123456,"$type":"asdf","Overs":10,"Balls":6,"TeamName":"TestTeam"}

Demo on .NET Fiddle: https://dotnetfiddle.net/a370Mv

edited May 30 '18 at 17:00

answered May 30 '18 at 16:37

Stringfellow

2,788
2
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But you are just creating static values here, this doesn't keep the original values. – DavidG May 30 '18 at 16:39
I am demonstrating that one can also update the `$id` and `$type` as well. – Stringfellow May 30 '18 at 16:47
But how would you keep the original values? This code just makes up new ones. – DavidG May 30 '18 at 16:47
1

This worked Stringfellow... I want to approve your code but I don't have the enough reputation(I am new user) so unable to approve your code. Your code worked for me. Thanks – Arul Joseph May 30 '18 at 17:12

galdin · Answer 3 · 2018-05-30T16:45:09.633

0

Use a concrete class. You'll need to annotate the fields with $ prefixes manually. Eg:

public class Example
{
    public string Field { get; set; }

    [JsonProperty("$type")]
    public string Type { get; set; }
}

Here's a working example.

In your case the class will look something like:

public class ObjName
{
    [JsonProperty("$id")]
    public string Id { get; set; }
    [JsonProperty("$type")]
    public string Type { get; set; }
    public int Overs { get; set; }
    public int Balls { get; set; }
    public string TeamName { get; set; }
}

(Just be mindful of the property case).

edited May 30 '18 at 16:45

answered May 30 '18 at 16:23

galdin

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7
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This won't work, the `$id` and `$type` properties are special values. – DavidG May 30 '18 at 16:25
@TypeIA Exactly, because they are special values used by JSON.Net which is why they occur at the start of the JSON. Look up `TypeNameHandling` (and are you now saying that you *didn't* try it out earlier...? :)) – DavidG May 30 '18 at 16:53
Always good to learn new things, and there's always more to learn! – DavidG May 30 '18 at 17:08
@DavidG Luckily the authors did the right thing and provided an option to control this behavior. Thus, it works how we tell it to work. I have posted a more thorough answer with this information. I will also do some pollution control and delete all my previous comments here. – TypeIA May 30 '18 at 17:28

JSON change value

3 Answers3