CODEWITHSUNDEEP
javajacksonobjectmapperfasterxml
Joe
Joe

Reputation: 9

com.fasterxml.jackson.databind.exc.UnrecognizedPropertyException: Unrecognized field "field" (class com.mapper.example.Parent)

I got a deserialization problem: Trying to convert class which extends parent class. Please find my class below,

public class Base {
    Parent parent;
    public Parent getParent() {
        return parent;
    }
    public void setParent(Parent parent) {
        this.parent = parent;
    }
}
-----------------
public class Parent {
    String name;
    String city;
    public String getName() {
        return name;
    }
    public void setName(String name) {
        this.name = name;
    }
    public String getCity() {
        return city;
    }
    public void setCity(String city) {
        this.city = city;
    }
}
-----------------
public class Child extends Parent {
    String field;
    public String getField() {
        return field;
    }
    public void setField(String field) {
        this.field = field;
    }
}

Main method:

public class TestDeserial {
    public static void main(String[] args) {
        Child c = new Child();
        c.setName("test1");
        c.setCity("Mumbai");
        c.setField("allow");
        Base b = new Base();
        b.setParent(c);
        // Convert object to json
        File file = new File("C:\\temp\\c.json");
        try {
            ObjectMapper mapper = new ObjectMapper();
            mapper.writeValue(file, b);
        } catch (IOException e) {
            e.printStackTrace();
        }
        
        // Convert json to object
        Base baseResult = null;
        try (InputStream in = new FileInputStream("C:\\temp\\c.json")) {
            ObjectMapper objMapper = new ObjectMapper();
            baseResult = objMapper.readValue(in, Base.class);
        } catch (Exception e1) {
            e1.printStackTrace();
        }
    }
}

Error: com.fasterxml.jackson.databind.exc.UnrecognizedPropertyException: Unrecognized field "field" (class com.mapper.example.Parent), not marked as ignorable (2 known properties: "name", "city"]) at [Source: (FileInputStream); line: 1, column: 52] (through reference chain: com.mapper.example.Base["parent"]->com.mapper.example.Parent["field"])

I don't want to add any annotation in class file.

Upvotes: -1

Views: 909

Answers (1)

jon hanson
jon hanson

Reputation: 9408

You need to tell Jackson to use deduction-based inference for determing the appropriate sub-class. Jackson will inspect the fields in the JSON and compare to the fields in the sub-classes.

If you don't want to add annotations to the original classes then you can use mix-ins, e.g.:

@JsonTypeInfo(use = JsonTypeInfo.Id.DEDUCTION)
@JsonSubTypes({ @JsonSubTypes.Type(Child.class) })
public abstract class ParentMixin {}

and then just register it with your ObjectMapper (the one used for deserialisation):

ObjectMapper mapper = new ObjectMapper();
mapper.addMixIn(Parent.class, ParentMixin.class);

Upvotes: 1

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