How to convert JSON to YAML using YamlDotNet

You don't actually need to deserialize JSON into strongly typed object you can convert JSON to YAML using dynamic Expando object as well. Here is a small example:-

var json = @"{
        'Name':'Peter',
        'Age':22,
        'CourseDet':{
                'CourseName':'CS',
                'CourseDescription':'Computer Science',
                },
        'Subjects':['Computer Languages','Operating Systems']
        }";

        var expConverter = new ExpandoObjectConverter();
        dynamic deserializedObject = JsonConvert.DeserializeObject<ExpandoObject>(json, expConverter);

        var serializer = new YamlDotNet.Serialization.Serializer();
        string yaml = serializer.Serialize(deserializedObject);

You can see a detailed explanation of both methods i.e. using strongly typed object and dynamic object here.


You can convert the JObject to a simpler object that YamlDotNet can serialize:

class Program
{
    static void Main(string[] args)
    {
        var json = "{\"swagger\":\"2.0\",\"info\":{\"title\":\"UberAPI\",\"description\":\"MoveyourappforwardwiththeUberAPI\",\"version\":\"1.0.0\"},\"host\":\"api.uber.com\",\"schemes\":[\"https\"],\"basePath\":\"/v1\",\"produces\":[\"application/json\"]}";
        var swaggerDocument = ConvertJTokenToObject(JsonConvert.DeserializeObject<JToken>(json));

        var serializer = new YamlDotNet.Serialization.Serializer();

        using (var writer = new StringWriter())
        {
            serializer.Serialize(writer, swaggerDocument);
            var yaml = writer.ToString();
            Console.WriteLine(yaml);
        }
    }

    static object ConvertJTokenToObject(JToken token)
    {
        if (token is JValue)
            return ((JValue)token).Value;
        if (token is JArray)
            return token.AsEnumerable().Select(ConvertJTokenToObject).ToList();
        if (token is JObject)
            return token.AsEnumerable().Cast<JProperty>().ToDictionary(x => x.Name, x => ConvertJTokenToObject(x.Value));
        throw new InvalidOperationException("Unexpected token: " + token);
    }
}

I think there is problem when json deserialization returns JObject. Looks like yaml serializer doesn't like it.

I used deserialization with specified type as you mentioned JsonConvert.DeserializeObject<SwaggerDocument>(json) and this is what I get

Swagger: 2.0
Info:
  Title: UberAPI
  Description: MoveyourappforwardwiththeUberAPI
  Version: 1.0.0
Host: api.uber.com
Schemes:
- https
BasePath: /v1
Produces:
- application/json

This is my whole code:

class Program
{
    static void Main(string[] args)
    {
        var json = "{\"Swagger\":\"2.0\",\"Info\":{\"Title\":\"UberAPI\",\"Description\":\"MoveyourappforwardwiththeUberAPI\",\"Version\":\"1.0.0\"},\"Host\":\"api.uber.com\",\"Schemes\":[\"https\"],\"BasePath\":\"/v1\",\"Produces\":[\"application/json\"]}";
        var swaggerDocument = JsonConvert.DeserializeObject<SwaggerDocument>(json);
        
        var serializer = new YamlDotNet.Serialization.Serializer();

        using (var writer = new StringWriter())
        {
            serializer.Serialize(writer, swaggerDocument);
            var yaml = writer.ToString();
            Console.WriteLine(yaml);
        }
    }
}

public class Info
{
    public string Title { get; set; }
    public string Description { get; set; }
    public string Version { get; set; }
}

public class SwaggerDocument
{
    public string Swagger { get; set; }
    public Info Info { get; set; }
    public string Host { get; set; }
    public List<string> Schemes { get; set; }
    public string BasePath { get; set; }
    public List<string> Produces { get; set; }
}

update

Two issues here.

When deserializing class with fields, by default, json.net won't take them into consideration when doing this job. For this purpose we have to customize the deserialization process by creating a custom contract resolver. We can easily do this by

var swaggerDocument = JsonConvert.DeserializeObject<SwaggerDocument>(json, new JsonSerializerSettings
{
    ContractResolver = new MyContractResolver()
});

public class MyContractResolver : DefaultContractResolver
{
    protected override IList<JsonProperty> CreateProperties(Type type, MemberSerialization memberSerialization)
    {
        var props = type.GetProperties(BindingFlags.Public | BindingFlags.Instance)
            .Select(p => base.CreateProperty(p, memberSerialization))
            .Union(type.GetFields(BindingFlags.Public | BindingFlags.Instance)
                .Select(f => base.CreateProperty(f, memberSerialization)))
            .ToList();
        props.ForEach(p => { p.Writable = true; p.Readable = true; });
        return props;
    }
}

There is second issue when we want to serialize class with fields: Values from fields won't be included into yaml result. I haven't figured out how to deal with this yet.

Do you have to use Swashbuckle.Swagger type or you can just create wrapper/decorator/DTO for this type?

I hope it helps you.