Use property of parent object to determine subclass when deserializing?

Question

children: [
      {
          o kind: "t3"
            data: {                 // ExampleNodeT3 class should be used for kind == t3
                + t3var1: "val1"
                + t3var2: true
            }
      }
      {
          o kind: "t4"                
            data: {                 // ExampleNodeT4 class should be used for kind == t4
                + t4var1: false
                + t4var2: 2346
            }
      }
] ... etc.


@JsonTypeInfo(use=Id.NAME, property="kind")
@JsonSubTypes({
@Type(value=ExampleNodeT3.class, name="t3"),
@Type(value=ExampleNodeT4.class, name="t4")})
public abstract class ExampleNode {
...
public void setData(ExampleNode data) {
    this.data = data;
}

When attempting to deserialize this with Jackson, the JsonTypeInfo hints fail when ExampleNode data is created because the "kind" property is associated with its parent and not visible. I have tried various variations of factory methods, and Jackson annotations, but because Jackson creates the ExampleNode object and passes it to setData() itself, I see no place to control what class of object is created.

Apparently, this functionality is not possible. The feature enhancement was requested here: http://jira.codehaus.org/browse/JACKSON-275 — phreakocious, Jan 03 '11 at 02:12
And the details of the problem are here: http://jackson-users.ning.com/forum/topics/how-to-use-a-property-of?commentId=5286555%3AComment%3A2679 — phreakocious, Jan 03 '11 at 02:12

score 8 · Accepted Answer · edited Jun 20 '20 at 09:12

Got to here from google , and found the solution. actually these days its possible due to the include=JsonTypeInfo.As.EXTERNAL_PROPERTY , example:

 public class Parent {

    @JsonProperty("type")
    public String type;

    @JsonProperty("data")
    @JsonInclude(Include.NON_NULL)
    public ChildBase ChildBase;
    
    public Parent() {
        medias = new HashMap<>();
    }
        
    @JsonTypeInfo(use=JsonTypeInfo.Id.CLASS, include=JsonTypeInfo.As.EXTERNAL_PROPERTY, property="type")
    @JsonTypeIdResolver(ChildBaseByParentTypeResolver.class)
    public void setChildBase(ChildBase ChildBase){
        this.ChildBase = ChildBase;
    }
}


@JsonInclude(JsonInclude.Include.NON_NULL)
@JsonIgnoreProperties(ignoreUnknown = true)
public class ChildBase {
      public String someStr;

}


@JsonInclude(JsonInclude.Include.NON_NULL)
@JsonIgnoreProperties(ignoreUnknown = true)
public class AggressiveChild extends ChildBase{
     public String someStr1;

}

@JsonInclude(JsonInclude.Include.NON_NULL)
@JsonIgnoreProperties(ignoreUnknown = true)
public class ChilledChild extends ChildBase{
     public String someStr1;

}


public class ChildBaseByParentTypeResolver extends TypeIdResolverBase {
    
    private JavaType superType;

    @Override
    public void init(JavaType baseType) {
        superType = baseType;
    }   
    
    @Override
    public Id getMechanism() {
        return Id.NAME;
    }
    
    @Override
    public JavaType typeFromId(DatabindContext context, String id) {
        Class<?> subType = ChildBase.class;     
        switch (id) {
        case "agressiveParent":        
            subType = AggressiveChild.class;
            break;
        case "chilledParent":        
            subType = ChilledChild.class;
            break;        
        }
        return context.constructSpecializedType(superType, subType);        
    }

    @Override
    public JavaType typeFromId(String directiveType) {
         throw new NotImplementedException();
    }

}

Interesting articles:

Polymorphism with jackson friend

Related stack overflow question

What if the property is a list? Like `public List childBases;`, in my test, it doesn't work. — mt eee, Dec 31 '19 at 06:09
in my case i didnt work with List BUT when you find answer, if you could comment here with it , it will be great. — Robocide, Dec 31 '19 at 10:03

score 1 · Answer 2 · answered Jan 03 '11 at 02:13

1

This functionality is not yet supported.

Feature enhancement request: http://jira.codehaus.org/browse/JACKSON-275

More details on the problem: http://jackson-users.ning.com/forum/topics/how-to-use-a-property-of?commentId=5286555%3AComment%3A2679

answered Jan 03 '11 at 02:13

phreakocious

123
2
8

score 1 · Answer 3 · answered Jan 04 '11 at 18:23

Correct -- use of "external type identifiers" (ones that are not contained within Object to deserialize, but as siblings) is not yet supported.

This could be implemented (as in, there is nothing fundamental preventing support from being added) -- beyond Jira issue mentioned that could help, a new feature request could be filed for directly requesting ability to use such type identifiers. This has actually been mentioned by multiple users; possibly since there are data formats (geoJSON?) that use such typing structure.

Use property of parent object to determine subclass when deserializing?

3 Answers3

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