Jackson databind enum case insensitive

JavaJsonSerializationEnumsJackson

Java Problem Overview


How can I deserialize JSON string that contains enum values that are case insensitive? (using Jackson Databind)

The JSON string:

[{"url": "foo", "type": "json"}]

and my Java POJO:

public static class Endpoint {
	
	public enum DataType {
		JSON, HTML
	}
	
	public String url;
	public DataType type;

	public Endpoint() {

	}

}

in this case,deserializing the JSON with "type":"json" would fail where as "type":"JSON" would work. But I want "json" to work as well for naming convention reasons.

Serializing the POJO also results in upper case "type":"JSON"

I thought of using @JsonCreator and @JsonGetter:

    @JsonCreator
	private Endpoint(@JsonProperty("name") String url, @JsonProperty("type") String type) {
		this.url = url;
		this.type = DataType.valueOf(type.toUpperCase());
	}

    //....
    @JsonGetter
	private String getType() {
		return type.name().toLowerCase();
	}

And it worked. But I was wondering whether there's a better solutuon because this looks like a hack to me.

I can also write a custom deserializer but I got many different POJOs that use enums and it would be hard to maintain.

Can anyone suggest a better way to serialize and deserialize enums with proper naming convention?

I don't want my enums in java to be lowercase!

Here is some test code that I used:

    String data = "[{\"url\":\"foo\", \"type\":\"json\"}]";
    Endpoint[] arr = new ObjectMapper().readValue(data, Endpoint[].class);
		System.out.println("POJO[]->" + Arrays.toString(arr));
		System.out.println("JSON ->" + new ObjectMapper().writeValueAsString(arr));

Java Solutions


Solution 1 - Java

Jackson 2.9

This is now very simple, using jackson-databind 2.9.0 and above

ObjectMapper objectMapper = new ObjectMapper();
objectMapper.enable(MapperFeature.ACCEPT_CASE_INSENSITIVE_ENUMS);

// objectMapper now deserializes enums in a case-insensitive manner

Full example with tests

import com.fasterxml.jackson.databind.MapperFeature;
import com.fasterxml.jackson.databind.ObjectMapper;

public class Main {

  private enum TestEnum { ONE }
  private static class TestObject { public TestEnum testEnum; }

  public static void main (String[] args) {
    ObjectMapper objectMapper = new ObjectMapper();
    objectMapper.enable(MapperFeature.ACCEPT_CASE_INSENSITIVE_ENUMS);

    try {
      TestObject uppercase = 
        objectMapper.readValue("{ \"testEnum\": \"ONE\" }", TestObject.class);
      TestObject lowercase = 
        objectMapper.readValue("{ \"testEnum\": \"one\" }", TestObject.class);
      TestObject mixedcase = 
        objectMapper.readValue("{ \"testEnum\": \"oNe\" }", TestObject.class);

      if (uppercase.testEnum != TestEnum.ONE) throw new Exception("cannot deserialize uppercase value");
      if (lowercase.testEnum != TestEnum.ONE) throw new Exception("cannot deserialize lowercase value");
      if (mixedcase.testEnum != TestEnum.ONE) throw new Exception("cannot deserialize mixedcase value");

      System.out.println("Success: all deserializations worked");
    } catch (Exception e) {
      e.printStackTrace();
    }
  }
}

Solution 2 - Java

I ran into this same issue in my project, we decided to build our enums with a string key and use @JsonValue and a static constructor for serialization and deserialization respectively.

public enum DataType {
    JSON("json"), 
    HTML("html");

    private String key;

    DataType(String key) {
        this.key = key;
    }

    @JsonCreator
    public static DataType fromString(String key) {
        return key == null
                ? null
                : DataType.valueOf(key.toUpperCase());
    }

    @JsonValue
    public String getKey() {
        return key;
    }
}

Solution 3 - Java

Since Jackson 2.6, you can simply do this:

    public enum DataType {
        @JsonProperty("json")
        JSON,
        @JsonProperty("html")
        HTML
    }

For a full example, see this gist.

Solution 4 - Java

In version 2.4.0 you can register a custom serializer for all the Enum types (link to the github issue). Also you can replace the standard Enum deserializer on your own that will be aware about the Enum type. Here is an example:

public class JacksonEnum {

    public static enum DataType {
        JSON, HTML
    }

    public static void main(String[] args) throws IOException {
        List<DataType> types = Arrays.asList(JSON, HTML);
        ObjectMapper mapper = new ObjectMapper();
        SimpleModule module = new SimpleModule();
        module.setDeserializerModifier(new BeanDeserializerModifier() {
            @Override
            public JsonDeserializer<Enum> modifyEnumDeserializer(DeserializationConfig config,
                                                              final JavaType type,
                                                              BeanDescription beanDesc,
                                                              final JsonDeserializer<?> deserializer) {
                return new JsonDeserializer<Enum>() {
                    @Override
                    public Enum deserialize(JsonParser jp, DeserializationContext ctxt) throws IOException {
                        Class<? extends Enum> rawClass = (Class<Enum<?>>) type.getRawClass();
                        return Enum.valueOf(rawClass, jp.getValueAsString().toUpperCase());
                    }
                };
            }
        });
        module.addSerializer(Enum.class, new StdSerializer<Enum>(Enum.class) {
            @Override
            public void serialize(Enum value, JsonGenerator jgen, SerializerProvider provider) throws IOException {
                jgen.writeString(value.name().toLowerCase());
            }
        });
        mapper.registerModule(module);
        String json = mapper.writeValueAsString(types);
        System.out.println(json);
        List<DataType> types2 = mapper.readValue(json, new TypeReference<List<DataType>>() {});
        System.out.println(types2);
    }
}

Output:

["json","html"]
[JSON, HTML]

Solution 5 - Java

If you're using Spring Boot 2.1.x with Jackson 2.9 you can simply use this application property:

spring.jackson.mapper.accept-case-insensitive-enums=true

Solution 6 - Java

I went for the solution of Sam B. but a simpler variant.

public enum Type {
	PIZZA, APPLE, PEAR, SOUP;

	@JsonCreator
	public static Type fromString(String key) {
		for(Type type : Type.values()) {
			if(type.name().equalsIgnoreCase(key)) {
				return type;
			}
		}
		return null;
	}
}

Solution 7 - Java

For those who tries to deserialize Enum ignoring case in GET parameters, enabling ACCEPT_CASE_INSENSITIVE_ENUMS will not do any good. It won't help because this option only works for body deserialization. Instead try this:

public class StringToEnumConverter implements Converter<String, Modes> {
    @Override
    public Modes convert(String from) {
        return Modes.valueOf(from.toUpperCase());
    }
}

and then

@Configuration
public class WebConfig implements WebMvcConfigurer {
 
    @Override
    public void addFormatters(FormatterRegistry registry) {
        registry.addConverter(new StringToEnumConverter());
    }
}

The answer and code samples are from here

Solution 8 - Java

To allow case insensitive deserialization of enums in jackson, simply add the below property to the application.properties file of your spring boot project.

spring.jackson.mapper.accept-case-insensitive-enums=true

If you have the yaml version of properties file, add below property to your application.yml file.

spring:
  jackson:
    mapper:
      accept-case-insensitive-enums: true

Solution 9 - Java

With apologies to @Konstantin Zyubin, his answer was close to what I needed - but I didn't understand it, so here's how I think it should go:

If you want to deserialize one enum type as case insensitive - i.e. you don't want to, or can't, modify the behavior of the entire application, you can create a custom deserializer just for one type - by sub-classing StdConverter and force Jackson to use it only on the relevant fields using the JsonDeserialize annotation.

Example:

public class ColorHolder {

  public enum Color {
    RED, GREEN, BLUE
  }

  public static final class ColorParser extends StdConverter<String, Color> {
    @Override
    public Color convert(String value) {
      return Arrays.stream(Color.values())
        .filter(e -> e.getName().equalsIgnoreCase(value.trim()))
        .findFirst()
        .orElseThrow(() -> new IllegalArgumentException("Invalid value '" + value + "'"));
	}
  }

  @JsonDeserialize(converter = ColorParser.class)
  Color color;
}

Solution 10 - Java

Problem is releated to com.fasterxml.jackson.databind.util.EnumResolver. it uses HashMap to hold enum values and HashMap doesn't support case insensitive keys.

in answers above, all chars should be uppercase or lowercase. but I fixed all (in)sensitive problems for enums with that:

https://gist.github.com/bhdrk/02307ba8066d26fa1537

CustomDeserializers.java

import com.fasterxml.jackson.databind.BeanDescription;
import com.fasterxml.jackson.databind.DeserializationConfig;
import com.fasterxml.jackson.databind.JsonDeserializer;
import com.fasterxml.jackson.databind.JsonMappingException;
import com.fasterxml.jackson.databind.deser.std.EnumDeserializer;
import com.fasterxml.jackson.databind.module.SimpleDeserializers;
import com.fasterxml.jackson.databind.util.EnumResolver;
 
import java.util.HashMap;
import java.util.Map;
 
 
public class CustomDeserializers extends SimpleDeserializers {
 
    @Override
    @SuppressWarnings("unchecked")
    public JsonDeserializer<?> findEnumDeserializer(Class<?> type, DeserializationConfig config, BeanDescription beanDesc) throws JsonMappingException {
        return createDeserializer((Class<Enum>) type);
    }
 
    private <T extends Enum<T>> JsonDeserializer<?> createDeserializer(Class<T> enumCls) {
        T[] enumValues = enumCls.getEnumConstants();
        HashMap<String, T> map = createEnumValuesMap(enumValues);
        return new EnumDeserializer(new EnumCaseInsensitiveResolver<T>(enumCls, enumValues, map));
    }
 
    private <T extends Enum<T>> HashMap<String, T> createEnumValuesMap(T[] enumValues) {
        HashMap<String, T> map = new HashMap<String, T>();
        // from last to first, so that in case of duplicate values, first wins
        for (int i = enumValues.length; --i >= 0; ) {
            T e = enumValues[i];
            map.put(e.toString(), e);
        }
        return map;
    }
 
    public static class EnumCaseInsensitiveResolver<T extends Enum<T>> extends EnumResolver<T> {
        protected EnumCaseInsensitiveResolver(Class<T> enumClass, T[] enums, HashMap<String, T> map) {
            super(enumClass, enums, map);
        }
 
        @Override
        public T findEnum(String key) {
            for (Map.Entry<String, T> entry : _enumsById.entrySet()) {
                if (entry.getKey().equalsIgnoreCase(key)) { // magic line <--
                    return entry.getValue();
                }
            }
            return null;
        }
    }
}

Usage:

import com.fasterxml.jackson.databind.ObjectMapper;
import com.fasterxml.jackson.databind.module.SimpleModule;
 

public class JSON {
 
    public static void main(String[] args) {
        SimpleModule enumModule = new SimpleModule();
        enumModule.setDeserializers(new CustomDeserializers());
 
        ObjectMapper mapper = new ObjectMapper();
        mapper.registerModule(enumModule);
    }
    
}

Solution 11 - Java

I used a modification of Iago Fernández and Paul solution .

I had an enum in my requestobject which needed to be case insensitive

@POST
public Response doSomePostAction(RequestObject object){
 //resource implementation
}



class RequestObject{
 //other params 
 MyEnumType myType;
 
 @JsonSetter
 public void setMyType(String type){
   myType = MyEnumType.valueOf(type.toUpperCase());
 }
 @JsonGetter
 public String getType(){
   return myType.toString();//this can change 
 }
}



Solution 12 - Java

Here's how I sometimes handle enums when I want to deserialize in a case-insensitive manner (building on the code posted in the question):

@JsonIgnore
public void setDataType(DataType dataType)
{
  type = dataType;
}

@JsonProperty
public void setDataType(String dataType)
{
  // Clean up/validate String however you want. I like
  // org.apache.commons.lang3.StringUtils.trimToEmpty
  String d = StringUtils.trimToEmpty(dataType).toUpperCase();
  setDataType(DataType.valueOf(d));
}

If the enum is non-trivial and thus in its own class I usually add a static parse method to handle lowercase Strings.

Solution 13 - Java

Deserialize enum with jackson is simple. When you want deserialize enum based in String need a constructor, a getter and a setter to your enum.Also class that use that enum must have a setter which receive DataType as param, not String:

public class Endpoint {

     public enum DataType {
        JSON("json"), HTML("html");
    
        private String type;
    
        @JsonValue
        public String getDataType(){
           return type;
        }
    
        @JsonSetter
        public void setDataType(String t){
           type = t.toLowerCase();
        }
     }

     public String url;
     public DataType type;

     public Endpoint() {
	
     }

     public void setType(DataType dataType){
        type = dataType;
     }

}

When you have your json, you can deserialize to Endpoint class using ObjectMapper of Jackson:

ObjectMapper mapper = new ObjectMapper();
mapper.enable(SerializationFeature.INDENT_OUTPUT);
try {
	Endpoint endpoint = mapper.readValue("{\"url\":\"foo\",\"type\":\"json\"}", Endpoint.class);
} catch (IOException e1) {
		// TODO Auto-generated catch block
	e1.printStackTrace();
}

Attributions

All content for this solution is sourced from the original question on Stackoverflow.

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Content TypeOriginal AuthorOriginal Content on Stackoverflow
Questiontom91136View Question on Stackoverflow
Solution 1 - JavadavnicwilView Answer on Stackoverflow
Solution 2 - JavaSam BerryView Answer on Stackoverflow
Solution 3 - JavaAndrew BickertonView Answer on Stackoverflow
Solution 4 - JavaAlexey GavrilovView Answer on Stackoverflow
Solution 5 - JavaharpresingView Answer on Stackoverflow
Solution 6 - JavalinquView Answer on Stackoverflow
Solution 7 - JavaKonstantin ZyubinView Answer on Stackoverflow
Solution 8 - JavaVivekView Answer on Stackoverflow
Solution 9 - JavaGussView Answer on Stackoverflow
Solution 10 - JavabhdrkView Answer on Stackoverflow
Solution 11 - Javatrooper31View Answer on Stackoverflow
Solution 12 - JavaPaulView Answer on Stackoverflow
Solution 13 - JavaIago FernándezView Answer on Stackoverflow