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Jackson Unrecognized field, not marked as ignorable

Jackson Unrecognized field, not marked as ignorable

1. Overview

In this article, we will learn to fix the Jackson UnrecognizedPropertyException: Unrecognized field, not marked as ignorable.

To learn more about the Jackson library, refer to our articles.

2. Jackson Unrecognized field, not marked as ignorable

Jackson ObjectMapper class supports both serialization and deserialization. Deserialization refers to converting JSON back to a Java object.

When your JSON contains additional fields than your Java object, then ObjectMapper throws the UnrecognizedPropertyException: Unrecognized field, not marked as ignorable during deserialization.

For example, the below Student class contains only the id field.

public class Student {
	
	public Student(long id) {
		this.id = id;
	}
	
	public Student() {
	}

	public long id;
	
	@Override
	public String toString() {
		return "Student [id=" + id + "]";
	}
}

Let’s assume that we had to deserialize the below JSON that contains id and name fields to the Student java object.

@Test
void unknownProperties_exception() throws JsonMappingException, JsonProcessingException {
	ObjectMapper mapper = new ObjectMapper();
	String json = "{\"id\": 1001, \"name\": \"Jackson\"}";
	Student student = mapper.readValue(json, Student.class);
	System.out.println(student);
}

Since the name field doesn’t exist in the Student class, the above deserialization results in the UnrecognizedPropertyException error.

com.fasterxml.jackson.databind.exc.UnrecognizedPropertyException: Unrecognized field "name" (class com.tedblob.jackson.examples.models.unknownproperties.Student), not marked as ignorable (one known property: "id"])

3. Solutions to fix Unrecognized field, not marked as ignorable

The possible reasons for this exception are:

  1. You don’t need all JSON fields in your Java class. So you skipped defining the unnecessary fields in the Java class. Check the next section for solutions.
  2. You declared the fields in Java class but they are private and doesn’t have any setter or getter methods defined for them. Be default, Jackson can deserialize only the fields that are public or contains setter or getters. To deserialize the private fields, see this article.

3.2. @JsonIgnoreProperties at the class level

You can use @JsonIgnoreProperties to ignore unknown properties at the class level. The ignoreUnknown property defines whether to ignore the unrecognized properties during the deserialization. If we specified it as true, then Jackson ignores unrecognized properties during deserialization.

@JsonIgnoreProperties(ignoreUnknown = true)
public class Student {
	
	public Student(long id) {
		this.id = id;
	}
	
	public Student() {
	}

	public long id;
	
	@Override
	public String toString() {
		return "Student [id=" + id + "]";
	}
}

3.3. Configure ObjectMapper globally to fix Unrecognized field, not marked as ignorable

You can ignore the unrecognized fields by configuring the ObjectMapper class:

mapper.configure(DeserializationFeature.FAIL_ON_UNKNOWN_PROPERTIES, false);

By declaring the FAIL_ON_UNKNOWN_PROPERTIES as false, the unrecognized properties are ignored.

@Test
void unknownProperties_configureMapper() throws JsonMappingException, JsonProcessingException {		
        ObjectMapper mapper = new ObjectMapper();
	mapper.configure(DeserializationFeature.FAIL_ON_UNKNOWN_PROPERTIES, false);
	String json = "{\"id\": 1001, \"name\": \"Jackson\"}";
	Student student = mapper.readValue(json, Student.class);
	System.out.println(student);
}

If you run the above test, then the JSON is deserialized properly to Student object.

Student [id=1001]

3.4. Ignore unrecognized properties globally using Jackson2ObjectMapperBuilder

If your project contains spring-web dependency, then you can use Jackson2ObjectMapperBuilder to build your ObjectMapper instance. By default, Jackson2ObjectMapperBuilder ignores the unrecognized properties and doesn’t throw any exception.

@Autowired
Jackson2ObjectMapperBuilder jsonBuilder;

@Test
void unknownProperties_JacksonBuilder() throws JsonMappingException, JsonProcessingException {
	ObjectMapper mapper = jsonBuilder.build();
	String json = "{\"id\": 1001, \"name\": \"Jackson\"}";
	Student student = mapper.readValue(json, Student.class);
	System.out.println(student);
}

4. Conclusion

To sum up, we have learned the different solutions to fix the Jackson Unrecognized field, not marked as ignorable.

2 thoughts on “Jackson Unrecognized field, not marked as ignorable”

  1. Pingback: Unrecognized Field Not Marked As Ignorable? 20 Most Correct Answers - Chambazone.com

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