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.JsonMappingException: 
    Can not construct instance of net.MyAbstractClass, 
    problem: abstract types either need to be mapped to concrete types, 
        have custom deserializer, or be instantiated with additional type information

I am having problems in the attribute:

@JsonTypeInfo(use = JsonTypeInfo.Id.NAME, include = JsonTypeInfo.As.WRAPPER_OBJECT, property = "@id")
@JsonSubTypes({ @JsonSubTypes.Type(value = MyAbstractClass.class, name = "MyAbstractClass") })
@ManyToOne
private MyAbstractClass object;

Could anyone help me?

Another discussion on a similar case is done at https://stackoverflow.com/a/32777371/873282. Especially interesting is the comment stating to use `@JsonTypeInfo(use = JsonTypeInfo.Id.CLASS, include = JsonTypeInfo.As.PROPERTY, property = "type")`. In that way, the JSON is a bit larger, but it is explicit, which type should be used. — koppor, Jun 27 '17 at 10:54

score 71 · Accepted Answer · edited Nov 12 '18 at 14:06

71

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

edited Nov 12 '18 at 14:06

thomas.mc.work

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answered Oct 05 '12 at 18:23

jlabedo

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Thanks for help, I resolved with: `JsonTypeInfo.Id.NAME, include = JsonTypeInfo.As.PROPERTY, property = "class")` the `property = "class"`, is coming in my JSON informing what child class will be used. – Danilo M. Oct 05 '12 at 20:45
1

Beware, allowing users to specify which class to use during deserialization might lead to vulnerabilities such as [this](http://fishbowl.pastiche.org/2015/11/09/java_serialization_bug/). – andres.riancho Dec 22 '15 at 16:24
2

Thanks for the reference: @JsonTypeInfo(use = JsonTypeInfo.Id.CLASS, include = JsonTypeInfo.As.PROPERTY, property = "@class") does the trick. Like the JPA table descriminator value. Thanks. – 99Sono Jul 04 '16 at 15:35

score 24 · Answer 2 · edited Dec 07 '21 at 15:48

24

I was having this issue using lombok's @Data and @Builder annotations alone, which I replaced with:

@Getter
@Setter
@NoArgsConstructor
@AllArgsConstructor
@Builder

And it solved my problem.

edited Dec 07 '21 at 15:48

cigien

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answered Dec 07 '21 at 14:25

Gabriel Andrade

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score 18 · Answer 3 · answered Jun 10 '20 at 09:03

18

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
    }

answered Jun 10 '20 at 09:03

Asif

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Worked for me here. Important to mention that `Vehicle` has a field `type`, which will create the correct object depending on the `name` defined of it. – finx Dec 22 '20 at 16:35
Does this work with resteasy? To me - not! – Alex Mi Dec 17 '21 at 07:36

score 6 · Answer 4 · answered Oct 06 '12 at 18:54

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).

score 5 · Answer 5 · answered Nov 23 '20 at 22:10

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.

Nice explanation. Had to put ```include = JsonTypeInfo.As.EXTERNAL_PROPERTY,``` to make it work for me. — maggle, May 01 '21 at 21:36

ZIA UR REHMAN · Answer 6 · 2020-01-29T10:17:43.053

4

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;
    }

}

edited Jan 29 '20 at 10:17

answered Jan 28 '20 at 19:09

ZIA UR REHMAN

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Maybe add an example for that could help people in the future – Sylhare Jan 28 '20 at 19:34

score 1 · Answer 7 · answered Aug 02 '21 at 13:51

1

Use @JsonDeserialize(contentAs= ConcreteCLass.class)

answered Aug 02 '21 at 13:51

Nuñito Calzada

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score 1 · Answer 8 · answered Jun 12 '23 at 15:15

1

Setting a default constructor helped to resolve

answered Jun 12 '23 at 15:15

sudha mosalikanti

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score 0 · Answer 9 · answered Jul 06 '21 at 18:20

Use lombok annotation.

@AllArgsConstructor
@NoArgsConstructor
@Builder
@JsonPropertyOrder({"name", "id"})
public class CreationUser {
    @Builder.Default
    @JsonProperty("name")
    String name = "";

    @Builder.Default
    @JsonProperty("id")
    String id = "";
}

score 0 · Answer 10 · answered Mar 03 '23 at 12:13

0

Using lombok I had the same issue, I have tried this to solve

@Getter
@Setter
@NoArgsConstructor
@AllArgsConstructor
@Builder

answered Mar 03 '23 at 12:13

Jean-Claude MBENG

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Cannot construct instance of - Jackson

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