Add property dynamically to a json string?

Question

I need to add some properties to a json string dynamically. Here is the code I'm using:

// set as empty json object
RequestMessage = "{}";
dynamic d = JsonConvert.DeserializeObject(RequestMessage);
d.Request = JsonConvert.SerializeObject(request);
d.RequestOptions = JsonConvert.SerializeObject(requestOptions);
RequestMessage = JsonConvert.SerializeObject(d);

This can add Request and RequestOptions to d, then serialize d back to json string.

It works fine if I know the properties' names, in this case, they are Request and RequestOptions.

Question is: is there a way to do this IF the property name is a variable? for example, something like:

private string GetJson(string name, object obj)
{
    // name is "Request"
    // object is request
    ......
    return RequestMessage;
}

Is it possible? *I'm using .net + newton json.

thanks

Why use dynamic at all? Just use a Dictionary<string, object> — maccettura
– maccettura, Commented Apr 6, 2020 at 15:39
Don't use dynamic. The concrete type Json.NET uses for a JSON object is JObject, so use that. See: How do you add a JToken to an JObject?, How do you Add or Update a JProperty Value in a JObject. A dictionary as mentioned above would also work. — dbc
– dbc, Commented Apr 6, 2020 at 15:40
the purpose is to add a property to the object dynamically (the object is converted from a json string), then convert it back to json string — urlreader
– urlreader, Commented Apr 6, 2020 at 15:41
You can add items to a Dictionary<string, object> or JObject in runtime. — dbc
– dbc, Commented Apr 6, 2020 at 15:43

Dmitry Ledentsov · Accepted Answer · 2020-04-06 15:43:19Z

2

Sure, you can add properties dynamically to a dynamic object:

var RequestMessage = "{}";
dynamic d = JsonConvert.DeserializeObject(RequestMessage);
d.Request = JsonConvert.SerializeObject(new { A = 42 });
d.RequestOptions = JsonConvert.SerializeObject(new { B = 22 });

var name = "SomeMore";
d[name]=11;

RequestMessage = JsonConvert.SerializeObject(d);

↓

{"Request":"{\"A\":42}","RequestOptions":"{\"B\":22}","SomeMore":11}

answered Apr 6, 2020 at 15:43

Dmitry Ledentsov

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3 Comments

maccettura Over a year ago

Why use dynamic instead of Dictionary<string, object> or JObject?

Dmitry Ledentsov Over a year ago

in this case d actually is a JObject

TC_Guy Over a year ago

MAJOR headache saver for me. Thanks!

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