JSON to YAML conversion (for .NET application)

Here are a few different ways to convert JSON to YAML in C#.

For JSON serialization/deserialization I have used Newtonsoft's Json.NET library and for YAML serialization/deserialization I have used YamlDotNet library.

• Using strongly typed C# object

  • Create the following classes to get the strongly typed object for JSON:-
  public class Student
  {
  
  public string Name { get; set; }
  public int Age { get; set; }
  public CourseDetails CourseDet { get; set; }
  public List<string> Subjects { get; set; }
  }
  
  
  public class CourseDetails
  {
  
  public string CourseName { get; set; }
  public string CourseDescription { get; set; }
  }
  
  
  • All set to go, the below code shows how to convert JSON to YAML using the "Student" class object as an intermediate :-
  var json = @"{
                  'Name':'Peter',
                  'Age':22,
                  'CourseDet':{
                          'CourseName':'CS',
                          'CourseDescription':'Computer Science',
                          },
                  'Subjects':['Computer Languages','Operating Systems']
                   }";
  
  var student = JsonConvert.DeserializeObject<Student>(json);
  
  var serializer = new YamlDotNet.Serialization.Serializer();
  
  using (var writer = new StringWriter())
  {
      serializer.Serialize(writer, student);
      var yaml = writer.ToString();
  }
  
  
  • The value of "yaml" variable after executing the above code segment is :-
  



• Using dynamic C# object

• With dynamic object you don't need to create any of those classes as above, all you need to do is use the below code directly :-
  
  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);
  
  The output of "yaml" variable here will be same as that for the strongly typed object.