How to Ignore Missing Properties During Jackson JSON Deserialization in Java

Question

How can I configure Jackson to ignore missing properties while deserializing JSON objects in Java?

class Person {
   String name;
   int age;
}

Answer

When deserializing JSON into Java objects using Jackson, you can encounter issues if certain properties in your JSON do not match the target class. By default, Jackson expects all defined properties to be present in the input JSON. However, you can configure Jackson to ignore any missing properties during deserialization, allowing for greater flexibility with JSON inputs.

import com.fasterxml.jackson.databind.ObjectMapper;
import com.fasterxml.jackson.databind.DeserializationFeature;
import com.fasterxml.jackson.annotation.JsonInclude;

@JsonInclude(JsonInclude.Include.NON_NULL)
class Person {
   String name;
   int age;
}

public class Main {
   public static void main(String[] args) throws Exception {
       ObjectMapper objectMapper = new ObjectMapper();
       objectMapper.configure(DeserializationFeature.FAIL_ON_UNKNOWN_PROPERTIES, false);
       String json = "{ 'name': 'John' }";
       Person person = objectMapper.readValue(json.replace("'", "\""), Person.class);
       System.out.println(person.name); // Output: John
       System.out.println(person.age); // Output: 0 (default int value for missing field)
   }
}

Causes

  • The target Java class has defined fields that are not present in the JSON input.
  • Jackson's default behavior is strict regarding field matching during deserialization.

Solutions

  • Use the `@JsonInclude` annotation for the class to specify how fields should be included during serialization/deserialization.
  • Set the `DeserializationFeature.FAIL_ON_UNKNOWN_PROPERTIES` feature to false.

Common Mistakes

Mistake: Forgetting to import necessary Jackson annotations and features.

Solution: Ensure you have imported `com.fasterxml.jackson.annotation.JsonInclude` and `com.fasterxml.jackson.databind.DeserializationFeature`.

Mistake: Using single quotes in JSON strings instead of double quotes.

Solution: Always use double quotes for JSON properties, or replace single quotes in strings before deserialization.

Helpers

  • Jackson JSON deserialization
  • ignore missing fields Jackson
  • Java JSON deserialization
  • Jackson annotations
  • handle missing properties JSON

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