如何在序列化特定类型时使JSON.Net序列化程序调用ToString()？

Question

我使用Newtonsoft.Json序列化程序将C#类转换为JSON.对于某些类,我不需要序列化程序将实例添加到单个属性,而只需要在对象上调用ToString,即

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

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

我应该怎么做才能将Person对象序列化为ToString()方法的结果？我可能有很多像这样的类,所以我不想最终得到一个特定于Person类的序列化器,我希望有一个可以适用于任何classe(通过属性我猜).

Answer 1

您可以使用自定义轻松完成此操作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();
    }
}

要使用转换器,请将需要序列化为类的任何类装饰为具有如下[JsonConverter]属性的字符串 :

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

这是一个演示转换器的演示:

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); 
    }
}

输出:

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

如果您还需要能够将字符串转换回对象,则可以ReadJson在转换器上实现该方法,以便查找public static Parse(string)方法并调用它.注意:请务必更改转换器的CanRead方法以返回true(或者只是CanRead完全删除重载),否则ReadJson将永远不会被调用.

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));
}

当然,为了使上述工作正常,您还需要确保Parse在要转换的每个类上实现一个合适的方法(如果它尚不存在).对于Person上面显示的示例类,该方法可能如下所示:

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;
}

往返演示:https://dotnetfiddle.net/fd4EG4

Answer 2

如果不打算大规模使用,有一种更快的方法,在下面的示例中,它是为RecordType属性完成的

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

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