Creating JSON in Windows 8 C# using IJsonValue: Cannot Implicitly convert type 'string' to 'Windows.Data.Json.IJsonValue'

Question

I am creating a Windows 8 App in C# using Visual Studio. I am trying to take a bunch of data input and create a JSON object from that data. Here is my preliminary test method that does not work:

        IJsonValue name = "Alex"; 
        JsonObject Character = new JsonObject();
        Character.Add("Name", name);

The error that I am getting is

Cannot Implicitly convert type 'string' to 'Windows.Data.Json.IJsonValue'

I looked up the documentation for IJsonValue but couldn't figure out how to create an instance of IJsonValue containing a string. So how do I store data in an IJsonValue to be added to a JsonObject?

IJsonValue is an interface, not a class (hence the "I" as prefix). The only thing you can assign to IJsonValue name is an object whose class implements the IJsonValue interface. — Flater
– Flater, Commented Feb 27, 2013 at 15:32
If your data is in a class, probably the easiest way to convert it to JSON would be to use the Newtonsoft JSON.NET library and call SerializeObject with your class instance... example here. — SWalters
– SWalters, Commented Feb 27, 2013 at 16:05

W.K.S · Accepted Answer · 2014-10-18 10:47:17Z

4

Here's how you would do it:

JsonObject character = new JsonObject();
character.Add("Name",JsonValue.CreateStringValue("Alex"));

answered Oct 18, 2014 at 10:47

W.K.S

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JP Alioto · Accepted Answer · 2013-02-27 16:46:19Z

3

The class JsonValue implements the IJsonValue interface. You can create an instance of the class and use it like this for example ...

JsonValue jsonValue = JsonValue.Parse("{\"Width\": 800, \"Height\": 600, \"Title\": \"View from 15th Floor\", \"IDs\": [116, 943, 234, 38793]}");
double width = jsonValue.GetObject().GetNamedNumber("Width");
double height = jsonValue.GetObject().GetNamedNumber("Height");
string title = jsonValue.GetObject().GetNamedString("Title");
JsonArray ids = jsonValue.GetObject().GetNamedArray("IDs");

Check out this for more examples.

answered Feb 27, 2013 at 16:46

JP Alioto

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1 Comment

Jerry Nixon Over a year ago

It's worth pointing out that handling JSON in this manner (almost like xpath) is a little brittle because there is little opportunity for typing and compile-time checks. Deserializing JSON into a strongly typed class is a more long-term and reliable approach that also gives you easier interaction with the data. It doesn't mean this GetNamedString approach does not work, it does. It's just a more typed and controlled approach that easier to use and less likely to break. :) Here's some info on it: stackoverflow.com/questions/10965829/…

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Creating JSON in Windows 8 C# using IJsonValue: Cannot Implicitly convert type 'string' to 'Windows.Data.Json.IJsonValue'

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