How to Deserialize JSON with Interface Properties in JSON.NET?

Casting Interfaces for Deserialization in JSON.NET

When working with JSON.NET for deserialization, a common issue arises when dealing with classes that contain interface-level properties. The error "Could not create an instance of type IThingy. Type is an interface or abstract class and cannot be instantiated" indicates that JSON.NET is unable to deserialize an interface.

To resolve this, utilize the solution proposed by @SamualDavis in a similar thread:

1. Include Concrete Classes as Constructor Parameters:

   When declaring the class that has interface properties, include the concrete classes as parameters to its constructor. This way, JSON.NET can identify the specific classes to be used during deserialization.

2. Constructor Example:

   public class Visit : IVisit
   {
       /// <summary>
       /// This constructor is required for the JSON deserializer to be able
       /// to identify concrete classes to use when deserializing the interface properties.
       /// </summary>
       public Visit(MyLocation location, Guest guest)
       {
           Location = location;
           Guest = guest;
       }
       public long VisitId { get; set; }
       public ILocation Location { get;  set; }
       public DateTime VisitDate { get; set; }
       public IGuest Guest { get; set; }
   }

By following this approach, JSON.NET will be able to successfully deserialize JSON objects into your C# objects, even if they contain interface-level properties.

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