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

Question

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.

Thanks for your reply. I tried the accepted answer, but with that solution we are missing the Child class member "field" in baseResult object. — Joe
– Joe, Commented Apr 4, 2023 at 9:13
If you don't want to add any annotation you can check the presence or not of the "field" property. If you want I can write a possible solution without annotations, but this could be resolved automatically with jackson annotations for polymorphism. — dariosicily
– dariosicily, Commented Apr 4, 2023 at 15:57

jon hanson · Accepted Answer · 2023-05-04 10:06:43Z

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);

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