datetime字段的反序列化问题

Question

反序列化JSON的最佳方法是什么

我有以下JSON

"_created" : { 
                "$dt": "2013-03-26T16:45:20Z" 
             }

和我想得到对象的字段，像这样- DataTime Created {get; set;}

json.net http://james.newtonking.com/projects/json-net.aspx专家提问

Answer 1

最简单的方法是创建jsonconverter

public class MongoDbDateTimeConverter : DateTimeConverterBase
    {
        public override object ReadJson(JsonReader reader, Type objectType, object existingValue, JsonSerializer serializer)
        {
            var jObject = JObject.Load(reader);
            return jObject["$dt"].Value<DateTime>();
        }
        public override void WriteJson(JsonWriter writer, object value, JsonSerializer serializer)
        {
            throw new NotImplementedException();
        }
    }

并与属性

一起使用

[JsonConverter(typeof(MongoDbDateTimeConverter))]
public DateTime Created { get; set; }

Answer 2

你可以使用NewtonSoft的JSON序列化器/反序列化器，对我来说工作得很好。

Nuget包

MediaTypeFormatter例子

将以下MediaTypeFormatter添加到GlobalConfiguration中，如下所示:

GlobalConfiguration.Configuration.Formatters.Remove(GlobalConfiguration.Configuration.Formatters.XmlFormatter);
GlobalConfiguration.Configuration.Formatters.Insert(0, new JsonFormatter());

MediaTypeFormatter:

public class JsonFormatter : MediaTypeFormatter
{
    private const string WesternEuropeStandardTime = "W. Europe Standard Time";
    private TimeZoneInfo timeZoneInfo;
    public JsonFormatter()
    {
        SupportedMediaTypes.Add(new System.Net.Http.Headers.MediaTypeHeaderValue("application/json"));
        this.timeZoneInfo = TimeZoneInfo.FindSystemTimeZoneById(WesternEuropeStandardTime);
    }
    public override bool CanReadType(Type type)
    {
        return true;
    }
    public override bool CanWriteType(Type type)
    {
        return true;
    }
    public override Task<object> ReadFromStreamAsync(Type type, Stream readStream, System.Net.Http.HttpContent content, IFormatterLogger formatterLogger)
    {
        Task<object> task = Task<object>.Factory.StartNew(() =>
        {
            JsonSerializerSettings settings = new JsonSerializerSettings()
            {
                NullValueHandling = NullValueHandling.Ignore,
            };
            StreamReader sr = new StreamReader(readStream);
            JsonTextReader jreader = new JsonTextReader(sr);
            JsonSerializer ser = new JsonSerializer();
            ser.Converters.Add(new DateTimeConverter(this.timeZoneInfo) { DateTimeFormat = "o" });
            return ser.Deserialize(jreader, type);
        });
        return task;
    }
    public override Task WriteToStreamAsync(Type type, object value, Stream writeStream, System.Net.Http.HttpContent content, System.Net.TransportContext transportContext)
    {
        Task task = Task.Factory.StartNew(() =>
        {
            JsonSerializerSettings settings = new JsonSerializerSettings()
            {
                NullValueHandling = NullValueHandling.Ignore,
            };
            string json = JsonConvert.SerializeObject(
                value, 
                Formatting.Indented,
                new JsonConverter[1] { new DateTimeConverter(this.timeZoneInfo) { DateTimeFormat = "o" } });
            byte[] buf = System.Text.Encoding.Default.GetBytes(json);
            writeStream.Write(buf, 0, buf.Length);
            writeStream.Flush();
        });
        return task;
    }
    private class DateTimeConverter : IsoDateTimeConverter
    {
        private TimeZoneInfo timeZoneInfo;
        public DateTimeConverter(TimeZoneInfo timeZoneInfo)
        {
            this.timeZoneInfo = timeZoneInfo;
        }
        public override void WriteJson(JsonWriter writer, object value, JsonSerializer serializer)
        {
            DateTime? date = value as DateTime?;
            if (date.HasValue && DateTime.MinValue != date.Value && DateTime.MaxValue != date.Value)
            {
                TimeSpan timeZoneOffset = this.timeZoneInfo.GetUtcOffset(date.Value);
                value = DateTime.SpecifyKind(date.Value - timeZoneOffset, DateTimeKind.Utc);
            }
            base.WriteJson(writer, value, serializer);
        }
        public override object ReadJson(JsonReader reader, Type objectType, object existingValue, JsonSerializer serializer)
        {
            object result = base.ReadJson(reader, objectType, existingValue, serializer);
            DateTime? date = result as DateTime?;
            if (date.HasValue && DateTime.MinValue != date.Value && DateTime.MaxValue != date.Value)
            {
                TimeSpan timeZoneOffset = this.timeZoneInfo.GetUtcOffset(date.Value);
                result = DateTime.SpecifyKind(date.Value + timeZoneOffset, DateTimeKind.Utc);
            }
            return result;
        }
    }
}

Answer 3

如果您使用json.NET

尝试serialize一个datetime字段或object组成的datetime字段，像这样:

JsonSerializerSettings microsoftDateFormatSettings = new JsonSerializerSettings
            {
                DateFormatHandling = DateFormatHandling.MicrosoftDateFormat
            };
string serializedObject= Newtonsoft.Json
                     .JsonConvert
                     .SerializeObject(data, microsoftDateFormatSettings);

如果使用JSON.NET完成序列化，它可以正常工作:)

然后你就可以成功反序列化它了

var myobject = Newtonsoft.Json.JsonConvert.DeserializeObject(serializedObject);

Answer 4

如果你问的是c#中的JSON解析器，那么这更多的是谷歌的事情。去那里找找看。有很多可用的

如果您正在询问如何将此字符串解析为DateTime，那么DateTime.TryParse()将帮助您。您可以通过提供字符串格式来帮助它，例如yyyy-MM-ddTHH:mm:ss。