Cannot construct instance of - Jackson

Question

I am using Jackson and I\'m having problems, when I try to deserialize an Object I get the following error:

com.fasterxml.jackson.databind.JsonMappingExcepti         


        

Answer 1

For me there was no default constructor defined for the POJOs I was trying to use. creating default constructor fixed it.

public class TeamCode {

    @Expose
    private String value;

    public String getValue() {
        return value;
    }

    **public TeamCode() {
    }**

    public TeamCode(String value) {
        this.value = value;
    }

    @Override
    public String toString() {
        return "TeamCode{" +
                "value='" + value + '\'' +
                '}';
    }

    public void setValue(String value) {
        this.value = value;
    }

}

Answer 2

You need to use a concrete class and not an Abstract class while deserializing. if the Abstract class has several implementations then, in that case, you can use it as below-

  @JsonTypeInfo( use = JsonTypeInfo.Id.NAME, include = JsonTypeInfo.As.PROPERTY, property = "type")
    @JsonSubTypes({ 
      @Type(value = Bike.class, name = "bike"), 
      @Type(value = Auto.class, name = "auto"), 
      @Type(value = Car.class, name = "car")
    })
    public abstract class Vehicle {
        // fields, constructors, getters, setters
    }

Answer 3

In your concrete example the problem is that you don't use this construct correctly:

@JsonSubTypes({ @JsonSubTypes.Type(value = MyAbstractClass.class, name = "MyAbstractClass") })

@JsonSubTypes.Type should contain the actual non-abstract subtypes of your abstract class.

Therefore if you have:

abstract class Parent and the concrete subclasses

Ch1 extends Parent and Ch2 extends Parent

Then your annotation should look like:

@JsonSubTypes({ 
          @JsonSubTypes.Type(value = Ch1.class, name = "ch1"),
          @JsonSubTypes.Type(value = Ch2.class, name = "ch2")
})

Here name should match the value of your 'discriminator':

@JsonTypeInfo(use = JsonTypeInfo.Id.NAME, 
include = JsonTypeInfo.As.WRAPPER_OBJECT, 
property = "type")

in the property field, here it is equal to type. So type will be the key and the value you set in name will be the value.

Therefore, when the json string comes if it has this form:

{
 "type": "ch1",
 "other":"fields"
}

Jackson will automatically convert this to a Ch1 class.

If you send this:

{
 "type": "ch2",
 "other":"fields"
}

You would get a Ch2 instance.

Answer 4

Your @JsonSubTypes declaration does not make sense: it needs to list implementation (sub-) classes, NOT the class itself (which would be pointless). So you need to modify that entry to list sub-class(es) there are; or use some other mechanism to register sub-classes (SimpleModule has something like addAbstractTypeMapping).

Answer 5

You cannot instantiate an abstract class, Jackson neither. You should give Jackson information on how to instantiate MyAbstractClass with a concrete type.

See this answer on stackoverflow: Jackson JSON library: how to instantiate a class that contains abstract fields

And maybe also see Jackson Polymorphic Deserialization