You can handle both formats (the JSON number representation and the masked string format) using a custom JsonConverter class like this. class DecimalConverter : JsonConverter { public override bool CanConvert(Type objectType) { return (objectType == typeof(decimal) || objectType == typeof(decimal?)); } public override object ReadJson(JsonReader reader, Type objectType, object existingValue, JsonSerializer serializer) { JToken token … Read more
Use this this site for representation: https://quicktype.io/csharp/ something like this may help you public class Item322A { public string prop1 { get; set; } public string prop2 { get; set; } public int prop3 { get; set; } public bool prop4 { get; set; } } public class Item2B { public string prop1 { get; … Read more
You should use JsonTypeInfo.As.EXTERNAL_PROPERTY instead of JsonTypeInfo.As.PROPERTY. In this scenario your Asset class should look like this: class Asset { @JsonTypeInfo( use = JsonTypeInfo.Id.NAME, include = JsonTypeInfo.As.EXTERNAL_PROPERTY, property = “type”) @JsonSubTypes({ @JsonSubTypes.Type(value = ImageAssetProperties.class, name = “image”), @JsonSubTypes.Type(value = DocumentAssetProperties.class, name = “document”) }) private AssetProperties properties; public AssetProperties getProperties() { return properties; } public … Read more
I’ve searched a lot and the best way I’ve found so far is on this article: Class to serialize package net.sghill.example; import net.sghill.example.UserDeserializer import net.sghill.example.UserSerializer import org.codehaus.jackson.map.annotate.JsonDeserialize; import org.codehaus.jackson.map.annotate.JsonSerialize; @JsonDeserialize(using = UserDeserializer.class) public class User { private ObjectId id; private String username; private String password; public User(ObjectId id, String username, String password) { this.id = … Read more
@DanielChapman gives a good explanation of serialVersionUID, but no solution. the solution is this: run the serialver program on all your old classes. put these serialVersionUID values in your current versions of the classes. as long as the current classes are serial compatible with the old versions, you should be fine. (note for future code: … Read more
Your question is, How can I serialize and deserialize a type with a string member that contains “raw” JSON, without escaping the JSON in the process? This can be done via a custom JsonConverter that reads and writes raw JSON using JsonWriter.WriteRawValue() and JRaw.Create(): public class RawConverter : JsonConverter { public override bool CanConvert(Type objectType) … Read more
You have to do it like this: import java.io.IOException; class Student implements java.io.Serializable { String name; String DOB; int id; Student(String naam, int idno, String dob) { name = naam; id = idno; DOB = dob; } private void writeObject(java.io.ObjectOutputStream stream) throws IOException { stream.writeObject(name); stream.writeInt(id); stream.writeObject(DOB); } private void readObject(java.io.ObjectInputStream stream) throws IOException, ClassNotFoundException … Read more
I am assuming you are not using Json.NET (Newtonsoft.Json NuGet package). If this the case, then you should try it. It has the following features: LINQ to JSON The JsonSerializer for quickly converting your .NET objects to JSON and back again Json.NET can optionally produce well formatted, indented JSON for debugging or display Attributes like … Read more
If you are using Json.NET, you can use JsonConvert.DeserializeObject<Dictionary<string, Dataset>>(json) and the keys of the dictionary will be nasdaq_imbalance, DXOpen IM, Float Shares
I am using like this in my code and it’s working fine below is a piece of code which you need to write using System.Web.Script.Serialization; JavaScriptSerializer oJS = new JavaScriptSerializer(); RootObject oRootObject = new RootObject(); oRootObject = oJS.Deserialize<RootObject>(Your JSon String);