How Can I Deserialize JSON Interface-Level Properties in JSON.NET?

JSON.NET Deserialization and Interface Properties: A Practical Solution

Deserializing JSON data into C# objects using JSON.NET can present difficulties when dealing with properties of interface types. A common error arises: JSON.NET can't directly instantiate interfaces.

The solution lies in leveraging concrete class parameters within the constructor of your class. This guides JSON.NET toward the correct concrete classes during deserialization.

Here's an example:

<code class="language-csharp">public class Visit : IVisit
{
    // Constructor crucial for JSON.NET to identify concrete types for deserialization.
    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; }
}</code>

By including MyLocation and Guest (concrete classes implementing ILocation and IGuest respectively) in the constructor, JSON.NET can successfully map the JSON data to the appropriate types, overcoming the interface instantiation problem. This approach allows you to retain the flexibility of interfaces while ensuring smooth deserialization.

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