De-serializing JSON to polymorphic object model using Spring and JsonTypeInfo annotation

Question

I have the following object model in my Spring MVC (v3.2.0.RELEASE) web application:

public class Order {
  private Payment payment;
}

@JsonTypeInfo(use = J         


        

Answer 1

Rashmin's answer worked, and I found an alternative way to avoid the com.fasterxml.jackson.databind.JsonMappingException: Could not resolve type id into a subtype of Blah issue without needing to use registerSubtypes. What you can do is add the following annotation to the parent class:

@JsonTypeInfo(use = JsonTypeInfo.Id.CLASS, include = JsonTypeInfo.As.PROPERTY, property = "type")

Note that the difference is JsonTypeInfo.Id.CLASS instead of JsonTypeInfo.Id.NAME. The downside is that the created JSON will contain the entire class name including the full namespace. The upside is that you don't have to worry about registering subtypes.

Answer 2

In my case I had added defaultImpl = SomeClass.class to @JsonTypeInfo and was trying to convert it SomeClass2.class

Answer 3

The method registerSubtypes() works!

@JsonTypeInfo(use=JsonTypeInfo.Id.NAME, include=JsonTypeInfo.As.PROPERTY, property="type")
public interface Geometry {
  //...
}

public class Point implements Geometry{
  //...
}
public class Polygon implements Geometry{
  //...
}
public class LineString implements Geometry{
  //...
}

GeoJson geojson= null;
ObjectMapper mapper = new ObjectMapper();
mapper.disable(DeserializationConfig.Feature.FAIL_ON_UNKNOWN_PROPERTIES);
mapper.registerSubtypes(Polygon.class,LineString.class,Point.class);

try {
    geojson=mapper.readValue(source, GeoJson.class);            
} catch (IOException e) {
    e.printStackTrace();
}

Note1: We use the Interface and the implementing classes. I fyou want jackson to de-serialize the classes as per their implementing classes, you have to register all of them using ObjectMapper's "registerSubtypes" method.

Note2: In addition you use, " @JsonTypeInfo(use=JsonTypeInfo.Id.NAME, include=JsonTypeInfo.As.PROPERTY, property="type")" as annotation with your Interface.

You can also define the order of properties when mapper writes a value of your POJO as a json.

This you can do using below annotation.

@JsonPropertyOrder({"type","crs","version","features"})
public class GeoJson {

    private String type="FeatureCollection";
    private List<Feature> features;
    private String version="1.0.0";
    private CRS crs = new CRS();
    ........
}

Hope this helps!

Answer 4

I had a similar issue while working on a dropwizard based service. I don't fully understand why things didn't work for me in the same way that the dropwizard code works, but I know why the code in the original post doesn't work. @JsonSubTypes wants an array of sub types, not a single value. So if you replace the line...

@JsonSubTypes.Type(name = "creditCardPayment", value = CreditCardPayment.class)

with...

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

I believe your code will work.

For those that are having this same error message pop up, you may be having an issue with the subtypes being discovered. Try adding a line like the one above or looking for issue with the discovery of the classes that have the @JsonTypeName tag in them.

Answer 5

Encountered the same error and used the equivalent of the below JSON (instead of CreditCardPayment used my class name) as the input for deserializer and it worked:

{
    "type": "CreditCardPayment", 
    ...
}

Answer 6

Try to register subtype using ObjectMapper.registerSubtypes instead of using annotations