将具有多种日期时间格式的 JSON 响应解析为 c# 对象

Question

This is what the JSON response is:

{
  "data": [
    {
      "someTicketNumber": "123456",
      "callInDateTimeUtc": "2020-09-09T20:16:26Z",
      "lastUpdatedDateTimeUtc": "2020-09-09T20:28:11Z",
      "availableDateLocalToSomeplace": "2020-09-10",
      "availableTimeLocalToSomeplace": "T10:30:00Z"
    }
  ]
}

A simple JsonConvert.DeserializeObject<T>(json) returns {9/10/2020 12:00:00 AM} for availableDateLocalToSomeplace and {9/11/2020 10:30:00 AM} for availableTimeLocalToSomeplace . I only want 9-10-2020 and 10:30:00 AM respectively. I can't put a generic converter settings like this:

JsonConvert.DeserializeObject<T>(json, new 
JsonSerializerSettings
{
    DateFormatString = "yyyy-MM-dd" OR "THH:mm:ssZ"
});

because I need 2 different format types. How do I accomplish this?

Answer 1

In this case you can do like this: Create a CustomDateTimeConverter class, and pass the format to it's constructor, like so

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

In the c# class,

public partial class Tickets
{
    [JsonProperty("data")]
    public TicketProps[] Data { get; set; }
}

public partial class TicketProps
{
    [JsonProperty("someTicketNumber")]
    public string SomeTicketNumber { get; set; }

    [JsonProperty("callInDateTimeUtc")]
    [JsonConverter(typeof(DateFormatConverter), "yyyy-MM-ddTHH:mm:ssZ")]
    public DateTime? CallInDateTimeUtc { get; set; }

    [JsonProperty("lastUpdatedDateTimeUtc")]
    [JsonConverter(typeof(DateFormatConverter), "yyyy-MM-ddTHH:mm:ssZ")]
    public DateTime? LastUpdatedDateTimeUtc { get; set; }

    [JsonProperty("availableDateLocalToSomeplace")]
    [JsonConverter(typeof(DateFormatConverter), "yyyy-MM-dd")]
    public DateTime?AvailableDateLocalToSomeplace { get; set; }

    [JsonProperty("availableTimeLocalToSomeplace")]
    [JsonConverter(typeof(DateFormatConverter), "THH:mm:ssZ")]
    public DateTime? AvailableTimeLocalToSomeplace { get; set; }
}

Now your simple JsonConvert.DeserializeObject<T>(json) will work just fine.