Dynamically deserialize from JSON

Question

I have some JSON here. The problem is it doesn't match the classes data types anymore. My question is; is it possible to deserialize JSON Dynamically? i.e. if I have entirely different JSON's can I deserialize them into two entirely different classes without first knowing what class I want to deserialize each into.

What do you mean by "adding another JSON type"? You mean adding another field? — Anastasios Selmani, Apr 09 '18 at 11:49
For example .The first json type is like the one in the top. And the second one is totally different JSON type with different fields. — Angelo Sanchez, Apr 09 '18 at 11:51
So two entirely different JSON's which would serialize into two entirely different classes? You say dynamically so do you not know what you want to serialize it into? — d219, Apr 09 '18 at 11:59
Possible duplicate of [Deserialize JSON into C# dynamic object?](https://stackoverflow.com/questions/3142495/deserialize-json-into-c-sharp-dynamic-object) — Anastasios Selmani, Apr 09 '18 at 13:14
Franctly if you just google your title or search in SO you will find the answer to your question immediately. https://stackoverflow.com/questions/3142495/deserialize-json-into-c-sharp-dynamic-object — Anastasios Selmani, Apr 09 '18 at 13:15

arslanaybars · Answer 1 · 2018-04-09T13:17:23.060

You can deserialize dynamic object with using newtonsoft

like bellowing code piece.

 dynamic dynamicObj = JsonConvert.DeserializeObject(jsonStr);
 string name = dynamicObj.data.code;

But in my personal preference is using strong type. I think its more convenience.

you can use quictype for generating c# classes from JSON object

quicktype generates strongly-typed models and serializers from JSON, JSON Schema, and GraphQL queries, making it a breeze to work with JSON type-safely in any programming language.

Hope the answer helps to you.

Dynamically deserialize from JSON

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