使用每种类型的不同格式在同一类中序列化多个DateTime属性

Question

我有一个具有两个DateTime属性的类.我需要使用不同的格式序列化每个属性.我该怎么做？我试过了:

JsonConvert.SerializeObject(obj, Formatting.None,
      new IsoDateTimeConverter {DateTimeFormat = "MM.dd.yyyy"});

此解决方案对我不起作用,因为它将日期格式应用于所有属性.有没有办法用不同的格式序列化每个DateTime属性？也许有一些属性？

Answer 1

处理这种情况的一种直接方法是子类化为IsoDateTimeConverter您需要的每种不同日期格式创建自定义日期转换器.例如:

class MonthDayYearDateConverter : IsoDateTimeConverter
{
    public MonthDayYearDateConverter()
    {
        DateTimeFormat = "MM.dd.yyyy";
    }
}

class LongDateConverter : IsoDateTimeConverter
{
    public LongDateConverter()
    {
        DateTimeFormat = "MMMM dd, yyyy";
    }
}

然后,您可以使用该[JsonConverter]属性来装饰DateTime需要自定义格式的任何类中的各个属性:

class Foo
{
    [JsonConverter(typeof(MonthDayYearDateConverter))]
    public DateTime Date1 { get; set; }

    [JsonConverter(typeof(LongDateConverter))]
    public DateTime Date2 { get; set; }

    // Use default formatting
    public DateTime Date3 { get; set; }
}

演示:

Foo foo = new Foo
{
    Date1 = DateTime.Now,
    Date2 = DateTime.Now,
    Date3 = DateTime.Now
};

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

输出:

{
  "Date1": "03.03.2014",
  "Date2": "March 03, 2014",
  "Date3": "2014-03-03T10:25:49.8885852-06:00"
}

Answer 2

NewtonSoft.Json 有一个有点难以理解的结构,你可以使用类似下面的自定义转换器来做你想要的:

[TestMethod]
public void Conversion()
{
    var obj = new DualDate()
    {
        DateOne = new DateTime(2013, 07, 25),
        DateTwo = new DateTime(2013, 07, 25)
    };
    Assert.AreEqual("{\"DateOne\":\"07.25.2013\",\"DateTwo\":\"2013-07-25T00:00:00\"}", 
        JsonConvert.SerializeObject(obj, Formatting.None, new DualDateJsonConverter()));
}

class DualDate
{
    public DateTime DateOne { get; set; }
    public DateTime DateTwo { get; set; }
}

class DualDateJsonConverter : JsonConverter
{

    public override void WriteJson(JsonWriter writer, object value, JsonSerializer serializer)
    {

        JObject result = new JObject();

        DualDate dd = (DualDate)value;

        result.Add("DateOne", JToken.FromObject(dd.DateOne.ToString("MM.dd.yyyy")));
        result.Add("DateTwo", JToken.FromObject(dd.DateTwo));
        result.WriteTo(writer);
    }

    // Other JsonConverterMethods
    public override bool CanConvert(Type objectType)
    {
        return objectType == typeof(DualDate);
    }

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

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

Answer 3

您可以创建一个继承IsoDateTimeConverter并在构造函数上传递格式的自定义日期类.在属性上,您可以指定与每个属性对应的格式.见下面的代码:

public class LoginResponse
{
    [JsonProperty("access_token")]
    public string AccessToken { get; set; }
    [JsonProperty("token_type")]
    public string TokenType { get; set; }
    [JsonProperty("expires_in")]
    public DateTime ExpiresIn { get; set; }
    [JsonProperty("userName")]
    public string Username { get; set; }
    [JsonConverter(typeof(CustomDateFormat), "EEE, dd MMM yyyy HH:mm:ss zzz")]
    [JsonProperty(".issued")]
    public DateTime Issued { get; set; }
    [JsonConverter(typeof(CustomDateFormat), "MMMM dd, yyyy")]
    [JsonProperty(".expires")]
    public DateTime Expires { get; set; }
}

public class CustomDateFormat : IsoDateTimeConverter
{
    public CustomDateFormat(string format)
    {
        DateTimeFormat = format;
    }
}