How can I add one more property using c# into existing json object inside Azure Functions? [duplicate]

Question

Inside the Azure function my input is ServiceBus queue properties

Code is to retrieve that all properties are -

using System.Net;
using Newtonsoft.Json;

public static async Task<HttpResponseMessage> Run(HttpRequestMessage req, TraceWriter log)
{

 string jsonContent = await req.Content.ReadAsStringAsync();

    return req.CreateResponse(HttpStatusCode.OK,jsonContent);
}

Output is -

[
    "{\"DeliveryCount\":\"1\",\MessageId\":\"bac52de2d23a487a9ed388f7313d93e5\"}"
]

I want to add one more property into this json object , how can I add it here in azure function so that I can return modified object like below -

[
    "{\"DeliveryCount\":\"1\",\MessageId\":\"bac52de2d23a487a9ed388f7313d93e5\",\"MyProperty\":\"TEST\"}"
]

serialise it to a type, add your property and deserialise again? — ADyson
– ADyson, Commented Jan 3, 2018 at 10:52
I dont have any type because i'm doing it into Azure Functions dynamically. — Neo
– Neo, Commented Jan 3, 2018 at 10:53
you could deserialise to dynamic. Doesn't have to be a concrete type. — ADyson
– ADyson, Commented Jan 3, 2018 at 10:54
dynamic x = JsonConvert.DeserializeObject<dynamic>(jsonContent); I would have thought. Then you can add a new property, and serialize it back again before returning it. — ADyson
– ADyson, Commented Jan 3, 2018 at 11:01
x.NewProperty = "y". That's the whole point about dynamic, it's, well...dynamic. — ADyson
– ADyson, Commented Jan 3, 2018 at 11:04

ADyson · Accepted Answer · 2018-01-03 11:24:19Z

1

I think you can do this quite easily, by de-serialising the JSON to an object, adding your new property and then serialising it back again. You don't even need a concrete type for this - dynamic should do the job for you.

For example:

public static async Task<HttpResponseMessage> Run(HttpRequestMessage req, TraceWriter log)
{
    string jsonContent = await req.Content.ReadAsStringAsync();
    dynamic obj = JsonConvert.DeserializeObject<dynamic>(jsonContent);
    obj.MyProperty = "TEST";
    string extendedJSON = JsonConvert.SerializeObject(obj);
    return req.CreateResponse(HttpStatusCode.OK, extendedJSON);
}

answered Jan 3, 2018 at 11:24

ADyson

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