CODEWITHSUNDEEP
javajsoncompiler-errorspolymorphismjackson
Jon Driscoll
Jon Driscoll

Reputation: 1483

Deserialize JSON with Jackson into Polymorphic Types - A Complete Example is giving me a compile error

I am attempting to work through a tutorial from Programmer Bruce that is supposed to allow the deserialization of polymorphic JSON.

The complete list can be found here Programmer Bruce tutorials (Great stuff btw)

I have worked through the first five with no problems but I have hit a snag on the last one (Example 6), which of course is the one I really need to get working.

I am getting the following error at compile time

The method readValue(JsonParser, Class) in the type ObjectMapper is not applicable for the arguments (ObjectNode, Class<capture#6-of ? extends Animal>)

and it's being caused by the chunk of code

  public Animal deserialize(  
      JsonParser jp, DeserializationContext ctxt)   
      throws IOException, JsonProcessingException  
  {  
    ObjectMapper mapper = (ObjectMapper) jp.getCodec();  
    ObjectNode root = (ObjectNode) mapper.readTree(jp);  
    Class<? extends Animal> animalClass = null;  
    Iterator<Entry<String, JsonNode>> elementsIterator =   
        root.getFields();  
    while (elementsIterator.hasNext())  
    {  
      Entry<String, JsonNode> element=elementsIterator.next();  
      String name = element.getKey();  
      if (registry.containsKey(name))  
      {  
        animalClass = registry.get(name);  
        break;  
      }  
    }  
    if (animalClass == null) return null;  
    return mapper.readValue(root, animalClass);
  }  
}

Specifically by the line

return mapper.readValue(root, animalClass);

Has anyone run into this before and if so, was there a solution?

Upvotes: 148

Views: 223733

Answers (7)

jbarrueta
jbarrueta

Reputation: 5175

As promised, I'm putting an example for how to use annotations to serialize/deserialize polymorphic objects, I based this example in the Animal class from the tutorial you were reading.

First of all your Animal class with the Json Annotations for the subclasses.

import com.fasterxml.jackson.annotation.JsonIgnoreProperties;
import com.fasterxml.jackson.annotation.JsonSubTypes;
import com.fasterxml.jackson.annotation.JsonTypeInfo;

@JsonIgnoreProperties(ignoreUnknown = true)
@JsonTypeInfo(use = JsonTypeInfo.Id.NAME, include = JsonTypeInfo.As.PROPERTY)
@JsonSubTypes({
    @JsonSubTypes.Type(value = Dog.class, name = "Dog"),

    @JsonSubTypes.Type(value = Cat.class, name = "Cat") }
)
public abstract class Animal {

    private String name;

    public String getName() {
        return name;
    }

    public void setName(String name) {
        this.name = name;
    }
    
}

Then your subclasses, Dog and Cat.

public class Dog extends Animal {

    private String breed;

    public Dog() {

    }

    public Dog(String name, String breed) {
        setName(name);
        setBreed(breed);
    }

    public String getBreed() {
        return breed;
    }

    public void setBreed(String breed) {
        this.breed = breed;
    }
}

public class Cat extends Animal {

    public String getFavoriteToy() {
        return favoriteToy;
    }

    public Cat() {}

    public Cat(String name, String favoriteToy) {
        setName(name);
        setFavoriteToy(favoriteToy);
    }

    public void setFavoriteToy(String favoriteToy) {
        this.favoriteToy = favoriteToy;
    }

    private String favoriteToy;

}

As you can see, there is nothing special for Cat and Dog, the only one that know about them is the abstract class Animal, so when deserializing, you'll target to Animal and the ObjectMapper will return the actual instance as you can see in the following test:

public class Test {

    public static void main(String[] args) {

        ObjectMapper objectMapper = new ObjectMapper();

        Animal myDog = new Dog("ruffus","english shepherd");

        Animal myCat = new Cat("goya", "mice");

        try {
            String dogJson = objectMapper.writeValueAsString(myDog);

            System.out.println(dogJson);

            Animal deserializedDog = objectMapper.readValue(dogJson, Animal.class);

            System.out.println("Deserialized dogJson Class: " + deserializedDog.getClass().getSimpleName());

            String catJson = objectMapper.writeValueAsString(myCat);

            Animal deseriliazedCat = objectMapper.readValue(catJson, Animal.class);

            System.out.println("Deserialized catJson Class: " + deseriliazedCat.getClass().getSimpleName());



        } catch (Exception e) {
            e.printStackTrace();
        }

    }
}

Output after running the Test class:

{"@type":"Dog","name":"ruffus","breed":"english shepherd"}

Deserialized dogJson Class: Dog

{"@type":"Cat","name":"goya","favoriteToy":"mice"}

Deserialized catJson Class: Cat

Upvotes: 221

Matthias M
Matthias M

Reputation: 14830

If the name of the existing property is not equal to name, you can use the annotation value EXISTING_PROPERTY

If the property name is for example type instead of name, you can use this annotation:

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

See also https://stackoverflow.com/a/62278471/1909531

Upvotes: 1

Xobotun
Xobotun

Reputation: 1371

Whereas @jbarrueta answer is perfect, in the 2.12 version of Jackson was introduced a new long-awaited type for the @JsonTypeInfo annotation, DEDUCTION.

It is useful for the cases when you have no way to change the incoming json or must not do so. I'd still recommend to use use = JsonTypeInfo.Id.NAME, as the new way may throw an exception in complex cases when it has no way to determine which subtype to use.

Now you can simply write

import com.fasterxml.jackson.annotation.JsonIgnoreProperties;
import com.fasterxml.jackson.annotation.JsonSubTypes;
import com.fasterxml.jackson.annotation.JsonTypeInfo;

@JsonIgnoreProperties(ignoreUnknown = true)
@JsonTypeInfo(use = JsonTypeInfo.Id.DEDUCTION)
@JsonSubTypes({
    @JsonSubTypes.Type(Dog.class),
    @JsonSubTypes.Type(Cat.class) }
)
public abstract class Animal {

    private String name;

    public String getName() {
        return name;
    }

    public void setName(String name) {
        this.name = name;
    }

}

And it will produce {"name":"ruffus", "breed":"english shepherd"} and {"name":"goya", "favoriteToy":"mice"}

Once again, it's safer to use NAME if some of the fields may be not present, like breed or favoriteToy.

Upvotes: 47

Dariusz
Dariusz

Reputation: 22241

Handling polymorphism is either model-bound or requires lots of code with various custom deserializers. I'm a co-author of a JSON Dynamic Deserialization Library that allows for model-independent json deserialization library. The solution to OP's problem can be found below. Note that the rules are declared in a very brief manner.

public class SOAnswer {
    @ToString @Getter @Setter
    @AllArgsConstructor @NoArgsConstructor
    public static abstract class Animal {
        private String name;    
    }

    @ToString(callSuper = true) @Getter @Setter
    @AllArgsConstructor @NoArgsConstructor
    public static class Dog extends Animal {
        private String breed;
    }

    @ToString(callSuper = true) @Getter @Setter
    @AllArgsConstructor @NoArgsConstructor
    public static class Cat extends Animal {
        private String favoriteToy;
    }
    
    
    public static void main(String[] args) {
        String json = "[{"
                + "    \"name\": \"pluto\","
                + "    \"breed\": \"dalmatian\""
                + "},{"
                + "    \"name\": \"whiskers\","
                + "    \"favoriteToy\": \"mouse\""
                + "}]";
        
        // create a deserializer instance
        DynamicObjectDeserializer deserializer = new DynamicObjectDeserializer();
        
        // runtime-configure deserialization rules; 
        // condition is bound to the existence of a field, but it could be any Predicate
        deserializer.addRule(DeserializationRuleFactory.newRule(1, 
                (e) -> e.getJsonNode().has("breed"),
                DeserializationActionFactory.objectToType(Dog.class)));
        
        deserializer.addRule(DeserializationRuleFactory.newRule(1, 
                (e) -> e.getJsonNode().has("favoriteToy"),
                DeserializationActionFactory.objectToType(Cat.class)));
        
        List<Animal> deserializedAnimals = deserializer.deserializeArray(json, Animal.class);
        
        for (Animal animal : deserializedAnimals) {
            System.out.println("Deserialized Animal Class: " + animal.getClass().getSimpleName()+";\t value: "+animal.toString());
        }
    }
}

Maven depenendency for pretius-jddl (check newest version at maven.org/jddl:

<dependency>
  <groupId>com.pretius</groupId>
  <artifactId>jddl</artifactId>
  <version>1.0.0</version>
</dependency>

Upvotes: 3

Marco
Marco

Reputation: 1575

You need only one line before the declaration of the class Animal for correct polymorphic serialization/deserialization:

@JsonTypeInfo(use = JsonTypeInfo.Id.CLASS, include = JsonTypeInfo.As.PROPERTY, property = "@class")
public abstract class Animal {
   ...
}

This line means: add a meta-property on serialization or read a meta-property on deserialization (include = JsonTypeInfo.As.PROPERTY) called "@class" (property = "@class") that holds the fully-qualified Java class name (use = JsonTypeInfo.Id.CLASS).

So, if you create a JSON directly (without serialization) remember to add the meta-property "@class" with the desired class name for correct deserialization.

More information here

Upvotes: 35

AmitW
AmitW

Reputation: 818

A simple way to enable polymorphic serialization / deserialization via Jackson library is to globally configure the Jackson object mapper (jackson.databind.ObjectMapper) to add information, such as the concrete class type, for certain kinds of classes, such as abstract classes.

To do that, just make sure your mapper is configured correctly. For example:

Option 1: Support polymorphic serialization / deserialization for abstract classes (and Object typed classes)

jacksonObjectMapper.enableDefaultTyping(
    ObjectMapper.DefaultTyping.OBJECT_AND_NON_CONCRETE);

Option 2: Support polymorphic serialization / deserialization for abstract classes (and Object typed classes), and arrays of those types.

jacksonObjectMapper.enableDefaultTyping(
    ObjectMapper.DefaultTyping.NON_CONCRETE_AND_ARRAYS);

Reference: https://github.com/FasterXML/jackson-docs/wiki/JacksonPolymorphicDeserialization

Upvotes: 7

ravi.zombie
ravi.zombie

Reputation: 1570

If using the fasterxml then,

these changes might be needed 

import com.fasterxml.jackson.core.JsonParser;
import com.fasterxml.jackson.core.JsonProcessingException;
import com.fasterxml.jackson.core.Version;
import com.fasterxml.jackson.databind.DeserializationContext;
import com.fasterxml.jackson.databind.JsonNode;
import com.fasterxml.jackson.databind.ObjectMapper;
import com.fasterxml.jackson.databind.deser.std.StdDeserializer;
import com.fasterxml.jackson.databind.module.SimpleModule;
import com.fasterxml.jackson.databind.node.ObjectNode;

in main method--

use

SimpleModule module =
  new SimpleModule("PolymorphicAnimalDeserializerModule");

instead of 

new SimpleModule("PolymorphicAnimalDeserializerModule",
      new Version(1, 0, 0, null));

and in Animal deserialize() function, make below changes

//Iterator<Entry<String, JsonNode>> elementsIterator =  root.getFields();
Iterator<Entry<String, JsonNode>> elementsIterator = root.fields();

//return mapper.readValue(root, animalClass);
return  mapper.convertValue(root, animalClass);

This works for fasterxml.jackson. If it still complains of the class fields. Use the same format as in the json for the field names (with "_" -underscore). as this
//mapper.setPropertyNamingStrategy(new CamelCaseNamingStrategy()); might not be supported.

abstract class Animal
{
  public String name;
}

class Dog extends Animal
{
  public String breed;
  public String leash_color;
}

class Cat extends Animal
{
  public String favorite_toy;
}

class Bird extends Animal
{
  public String wing_span;
  public String preferred_food;
}

Upvotes: -1

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