How to make JSON.Net serializer to call ToString() when serializing a particular type?

I am using Newtonsoft.Json serializer to convert C# classes to JSON. For some classes I don't need the serializer to an instance to individual properties, but instead just call ToString on the object, i.e.

public class Person
{
   public string FirstName { get; set; }
   public string LastName { get; set; }

   public override string ToString() { return string.Format("{0} {1}", FirstName, LastName ); }
}

What should I do to get the Person object serialized as the result of its ToString() method? I may have many classes like this, so I don't want to end up with a serializer specific for Person class, I want to have one than can be applicable to any classe (via attribute I guess).


You can do this easily with a custom JsonConverter:

public class ToStringJsonConverter : JsonConverter
{
    public override bool CanConvert(Type objectType)
    {
        return true;
    }

    public override void WriteJson(JsonWriter writer, object value, JsonSerializer serializer)
    {
        writer.WriteValue(value.ToString());
    }

    public override bool CanRead
    {
        get { return false; }
    }

    public override object ReadJson(JsonReader reader, Type objectType, object existingValue, JsonSerializer serializer)
    {
        throw new NotImplementedException();
    }
}

To use the converter, decorate any classes that need to be serialized as string with a [JsonConverter] attribute like this:

[JsonConverter(typeof(ToStringJsonConverter))]
public class Person
{
    ...
}

Here is a demo showing the converter in action:

class Program
{
    static void Main(string[] args)
    {
        Company company = new Company
        {
            CompanyName = "Initrode",
            Boss = new Person { FirstName = "Head", LastName = "Honcho" },
            Employees = new List<Person>
            {
                new Person { FirstName = "Joe", LastName = "Schmoe" },
                new Person { FirstName = "John", LastName = "Doe" }
            }
        };

        string json = JsonConvert.SerializeObject(company, Formatting.Indented);
        Console.WriteLine(json);
    }
}

public class Company
{
    public string CompanyName { get; set; }
    public Person Boss { get; set; }
    public List<Person> Employees { get; set; }
}

[JsonConverter(typeof(ToStringJsonConverter))]
public class Person
{
    public string FirstName { get; set; }
    public string LastName { get; set; }

    public override string ToString() 
    { 
        return string.Format("{0} {1}", FirstName, LastName); 
    }
}

Output:

{
  "CompanyName": "Initrode",
  "Boss": "Head Honcho",
  "Employees": [
    "Joe Schmoe",
    "John Doe"
  ]
}

If you also need to be able to convert from string back to an object, you can implement the ReadJson method on the converter such that it looks for a public static Parse(string) method and calls it. Note: be sure to change the converter's CanRead method to return true (or just delete the CanRead overload altogether), otherwise ReadJson will never be called.

public override object ReadJson(JsonReader reader, Type objectType, object existingValue, JsonSerializer serializer)
{
    MethodInfo parse = objectType.GetMethod("Parse", new Type[] { typeof(string) });
    if (parse != null && parse.IsStatic && parse.ReturnType == objectType)
    {
        return parse.Invoke(null, new object[] { (string)reader.Value });
    }

    throw new JsonException(string.Format(
        "The {0} type does not have a public static Parse(string) method that returns a {0}.", 
        objectType.Name));
}

Of course, for the above to work, you will also need to make sure to implement a suitable Parse method on each class you're converting, if it doesn't already exist. For our example Person class shown above, that method might look something like this:

public static Person Parse(string s)
{
    if (string.IsNullOrWhiteSpace(s))
        throw new ArgumentException("s cannot be null or empty", "s");

    string[] parts = s.Split(new char[] { ' ' }, 2);
    Person p = new Person { FirstName = parts[0] };
    if (parts.Length > 1)
        p.LastName = parts[1];

    return p;
}

Round-trip demo: https://dotnetfiddle.net/fd4EG4


There is a quicker way of doing this if it is not intended to be used on a large scale, in the example below, it is done for the RecordType property

[JsonIgnore]
public RecordType RecType { get; set; }

[JsonProperty(PropertyName = "RecordType")]
private string RecordTypeString => RecType.ToString();

You can simply try Newtonsoft's JSON builder library and Serilaize the object of type Person using such code:

Dictionary<string, object> collection = new Dictionary<string, object>()
    {
      {"First", new Person(<add FirstName as constructor>)},
      {"Second", new Person(<add LastName as constructor>)},

    };
string json = JsonConvert.SerializeObject(collection, Formatting.Indented, new JsonSerializerSettings
  {
    TypeNameHandling = TypeNameHandling.All,
    TypeNameAssemblyFormat = FormatterAssemblyStyle.Simple
  });