JSON Date and DateTime serialisation in c# & newtonsoft

Question

We are sending JSON to an API defined by swagger that some properties are DateTime in the format yyyy-MM-ddThh:mm:ss.000Z (the milliseconds must be 3 digits or it fails validation at the endpoint) and some are Date (no time) properties.

I have seen many messages saying use the formatters like this:

var jsonSettings = new JsonSerializerSettings();
jsonSettings.DateFormatString = "yyyy-MM-ddThh:mm:ss.000Z"; //try .fffZ too
var jsonObject= Newtonsoft.Json.JsonConvert.DeserializeObject<OurSwaggerObject>(json , setting);

but this does not convert the DateTimes into the correct format, and how does C# deal with a Date only type? It always seems to serialise as DateTime.MinValue()

Here is an example:

Someone sends me json as string but the the dates and datetimes in the incorrect format to be sent to the endpoint. I was hoping that the swagger class and json deserialisation would format them but it is not.

This is the swagger generated class

 public class OurSwaggerObject
    {
        [Newtonsoft.Json.JsonProperty("dateTimeField", Required = Newtonsoft.Json.Required.Always)]
        [System.ComponentModel.DataAnnotations.Required]
        [System.ComponentModel.DataAnnotations.RegularExpression(@"^\d{4}-\d\d-\d\dT\d\d:\d\d:\d\d\.\d{3}Z$")]
        public DateTime dateTimeField { get; set; }

        [Newtonsoft.Json.JsonProperty("dateField", Required = Newtonsoft.Json.Required.Always)]
        [System.ComponentModel.DataAnnotations.Required]
        [System.ComponentModel.DataAnnotations.RegularExpression(@"^\d{4}-\d\d-\d\d$")]
        public DateTime dateField { get; set; }
    }

So I try and coerce the json to be correct but I'm doing it wrong or something is missing

string json = @"{ 'dateTimeField': '1995-04-07T00:00:00',
                          'dateField': '1995-04-07T00:00:00'
                           }";

        /* The json we need to satisfy the swagger endpoint is:

          { 'dateTimeField': '1995-04-07T00:00:00.000Z',
            'dateField': '1995-04-07'
                           }              
          */

        OurSwaggerObject deserialisedIntoObject = Newtonsoft.Json.JsonConvert.DeserializeObject<OurSwaggerObject>(json);

        string serialisedToString = Newtonsoft.Json.JsonConvert.SerializeObject(deserialisedIntoObject);
        //serialisedToString= "{\"dateTimeField\":\"1995-04-07T00:00:00\",\"dateField\":\"1995-04-07T00:00:00\"}"

        var jsonSettings = new JsonSerializerSettings();
        jsonSettings.DateFormatString = "yyyy-MM-ddThh:mm:ss.fffZ"; //this won't help much for the 'date' only field!
        deserialisedIntoObject = Newtonsoft.Json.JsonConvert.DeserializeObject<OurSwaggerObject>(json,jsonSettings);
        serialisedToString = Newtonsoft.Json.JsonConvert.SerializeObject(deserialisedIntoObject, jsonSettings);
        //serialisedToString="{\"dateTimeField\":\"1995-04-07T00:00:00\",\"dateField\":\"1995-04-07T00:00:00\"}"

Answer 1

As I mentioned in a comment, there is no standard date representation in JSON. The ISO8601 is the de-facto standard, ie most people started using this some years ago. ISO8601 does not require milliseconds. If the other endpoint requires them, it's violating the defacto standard.

Json.NET uses IOS8601 since version 4.5. The current one is 10.0.3. The following code :

JsonConvert.SerializeObject(DateTime.Now)

returns

"2017-09-08T19:01:55.714942+03:00"

On my machine. Notice the timezone offset. That's also part of the standard. Z means UTC.

You can specify your own time format, provided it's the correct one. In this case, it should be yyyy-MM-ddTHH:mm:ss.fffZ. Notice the fff for milliseconds and HH for 24-hour.

The following code

var settings=new JsonSerializerSettings{DateFormatString ="yyyy-MM-ddTHH:mm:ss.fffZ"};
var json=JsonConvert.SerializeObject(DateTime.Now,settings);

returns

"2017-09-08T19:04:14.480Z"

The format string does not force a timezone translation. You can tell Json.NET to treat the time as Local or Utc through the DateTimeZoneHandling setting :

var settings=new JsonSerializerSettings{
                              DateFormatString ="yyyy-MM-ddTH:mm:ss.fffZ",
                              DateTimeZoneHandling=DateTimeZoneHandling.Utc};
var json=JsonConvert.SerializeObject(DateTime.Now,settings);

Returns :

"2017-09-08T16:08:19.290Z"

UPDATE

As Matt Johnson explains, Z is just a literal, while K generates either Z or an offset, depending on the DateTimeZoneHandling setting.

The format string yyyy-MM-ddTH:mm:ss.fffK with DateTimeZoneHandling.Utc :

var settings=new JsonSerializerSettings{
                              DateFormatString ="yyyy-MM-ddTH:mm:ss.fffK",
                              DateTimeZoneHandling=DateTimeZoneHandling.Utc};
var json=JsonConvert.SerializeObject(DateTime.Now,settings);

Will return :

2017-09-11T9:10:08.293Z

Changing to DateTimeZoneHandling.Utc will return

2017-09-11T12:15:12.862+03:00

Which, by the way is the default behaviour of Json.NET, apart from the forced millisecond precision.

Finally, .NET doesn't have a Date-only type yet. DateTime is used for both dates and date+time values. You can get the date part of a DateTime with the DateTime.Date property. You can retrieve the current date with DateTime.Today.

Time of day is represented by the Timespan type. You can extract the time of day from a DateTime value with DateTime.TimeOfDay. Timespan isn't strictly a time-of-day type as it can represent more than 24 hours.

What was that yet?

Support for explicit Date, TimeOfDay is comming through the CoreFX Lab project. This contains "experimental" features that are extremely likely to appear in the .NET Runtime like UTF8 support, Date, String, Channles. Some of these already appear as separate NuGet packages.

One can use the System.Time classes already, either by copying the code or adding them through the experimental NuGet source

Answer 2

Get current universaltime to json date time format and vice versa:

DateTime currentDateTime = DateTime.Now.ToUniversalTime();
var jsonDateTime = GetJSONFromUserDateTime(currentDateTime);
DateTime getDateTime = GetUserDateTimeFromJSON(jsonDateTime);

Here are both methods:

/// <summary>
/// Convert UserDateTime({9/7/2018 8:37:20 AM}) to JSON datetime(1536309440373) format
/// </summary>
/// <param name="givenDateTime"></param>
/// <returns></returns>
public static string GetJSONFromUserDateTime(DateTime givenDateTime)
{
    string jsonDateTime = string.Empty;
    if (givenDateTime != null)
    {
        JsonSerializerSettings microsoftDateFormatSettings = new JsonSerializerSettings
        {
            DateFormatHandling = DateFormatHandling.MicrosoftDateFormat
        };
        jsonDateTime = JsonConvert.SerializeObject(givenDateTime, microsoftDateFormatSettings);
        jsonDateTime = jsonDateTime.Replace("\"\\/Date(", "").Replace(")\\/\"", "");
    }
    return jsonDateTime;
}

/// <summary>
/// Convert JSON datetime(1536309440373) to user datetime({9/7/2018 8:37:20 AM})
/// </summary>
/// <param name="jsonDateTime"></param>
/// <returns></returns>
public static dynamic GetUserDateTimeFromJSON(string jsonDateTime)
{
    dynamic userDateTime = null;
    if (!string.IsNullOrEmpty(jsonDateTime))
    {
        JsonSerializerSettings microsoftDateFormatSettings = new JsonSerializerSettings
        {
            DateFormatHandling = DateFormatHandling.MicrosoftDateFormat
        };
        userDateTime = JsonConvert.DeserializeObject("\"\\/Date(" + jsonDateTime + ")\\/\"", microsoftDateFormatSettings);
    }
    return userDateTime;
}

Answer 3

If, like Willie Esteche you wish to convert a number that is too low, you have probably got a Unix Date Time value. His solution does indeed work, but as of .Net Framework 4.6 there is an easier way to do things.

Given a value of 1500013000, first you convert this to a DateTimeOffset with the FromUnixTimeSeconds() method, then simply grab the DateTime component.

DateTime dt = DateTimeOffset.FromUnixTimeSeconds(1500013000).UtcDateTime;

Conversion back (assuming UTC) is performed like so:

long Udt = new DateTimeOffset(dt,TimeSpan.Zero).ToUnixTimeSeconds();