CODEWITHSUNDEEP
c#.netdatetimejson.net
Felipe Deveza
Felipe Deveza

Reputation: 1969

JsonConvert.DeserializeObject with two types of datetime format

I need to expect both: dd/MM/yyyy and dd/MM/yyyy HH:mm:ss

var dateTimeConverter = new IsoDateTimeConverter { DateTimeFormat = "dd/MM/yyyy HH:mm:ss" };
var resultado = JsonConvert.DeserializeObject<MyObject>(json, dateTimeConverter);

With this code, i received the following error message:

System.FormatException: 'String was not recognized as a valid DateTime.'

Upvotes: 1

Views: 4311

Answers (3)

Alexander
Alexander

Reputation: 607

You can write your own JsonConverter:

class DataObject
{
    public DateTime CreatedDate { get; set; }
}

class CustomJsonConverter : JsonConverter
{
        public override object ReadJson(JsonReader reader, Type objectType, object existingValue, JsonSerializer serializer)
        {
            var obj = new DataObject();

            reader.Read();
            var prop = obj.GetType().GetProperty("CreatedDate");

            reader.Read();
            var strDate = (string)reader.Value;
            DateTime date;
            if (DateTime.TryParseExact(strDate, "dd/MM/yyyy HH:mm:ss", CultureInfo.InvariantCulture, DateTimeStyles.None, out date))
                prop.SetValue(obj, date);
            if (DateTime.TryParseExact(strDate, "dd/MM/yyyy", CultureInfo.InvariantCulture, DateTimeStyles.None, out date))
                prop.SetValue(obj, date);

            return obj;
        }
}

Upvotes: 1

Matheus Lacerda
Matheus Lacerda

Reputation: 6007

Maybe you could subclass IsoDateTimeConverter like this:

class Format1 : IsoDateTimeConverter
{
    public Format1()
    {
        DateTimeFormat = @"dd/MM/yyyy";
    }
}

class Format2 : IsoDateTimeConverter
{
    public Format2()
    {
        DateTimeFormat = @"dd/MM/yyyy HH:mm:ss";
    }
}

And then you just add JsonConverter attributes to your model, like this:

class Entity
{
    [JsonConverter(typeof(Format1))]
    public DateTime Date1 { get; set; }

Upvotes: 0

Matt Johnson-Pint
Matt Johnson-Pint

Reputation: 241525

You don't need to specify any converter at all. By default, both of those formats will work.

class Foo
{
    public DateTime DateTime { get; set; }
}

// This Just Works.

string json1 = "{ \"DateTime\" : \"12/31/2017\" }";
string json2 = "{ \"DateTime\" : \"12/31/2017 23:59:59\" }";

var o1 = JsonConvert.DeserializeObject<Foo>(json1);
var o2 = JsonConvert.DeserializeObject<Foo>(json2);

Upvotes: 4

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