我喜欢公认的答案。但是，我想稍微改进一下(考虑到现在有高于版本6的Java可用)。

例子:

    public enum Operation {
        EQUAL("eq"),
        NOT_EQUAL("ne"),
        LESS_THAN("lt"),
        GREATER_THAN("gt");

        private final String value;

        Operation(String value) {
            this.value = value;
        }

        @JsonValue
        public String getValue() {
            return value;
        }

        @JsonCreator
        public static Operation forValue(String value) {
            return Arrays.stream(Operation.values())
                .filter(op -> op.getValue().equals(value))
                .findFirst()
                .orElseThrow(); // depending on requirements: can be .orElse(null);
        }
    }

我一直在寻找枚举序列化的解决方案，我终于做出了一个解决方案。

https://github.com/sirgilligan/EnumerationSerialization

https://digerati-illuminatus.blogspot.com/2022/10/java-enum-generic-serializer-and.html

它使用了一个新的注释和两个新类EnumerationSerializer和EnumerationDeserializer。您可以子类化EnumerationDeserializer，并创建一个设置枚举class的类(典型方法)，或者您可以注释枚举，并且不必拥有EnumerationDeserializer的子类。

@JsonSerialize(using = EnumerationSerializer.class)
@JsonDeserialize(using = EnumerationDeserializer.class)
@EnumJson(serializeProjection = Projection.NAME, deserializationClass = RGB.class)
enum RGB {
    RED,
    GREEN,
    BLUE
}

注意ContextualDeserializer的实现如何从注释中提取类。

https://github.com/sirgilligan/EnumerationSerialization/blob/main/src/main/java/org/example/EnumerationDeserializer.java

这里有很多很好的代码，可以提供一些见解。

对于你的具体问题，你可以这样做:

@JsonSerialize(using = EnumerationSerializer.class)
@JsonDeserialize(using = EnumerationDeserializer.class)
@EnumJson(serializeProjection = Projection.NAME, deserializationClass = Event.class)
public enum Event {
    FORGOT_PASSWORD("forgot password");

    //This annotation is optional because the code looks for value or alias.
    @EnumJson(serializeProjection = Projection.VALUE)
    private final String value;

    private Event(final String description) {
        this.value = description;
    }

}

或者你可以这样做:

@JsonSerialize(using = EnumerationSerializer.class)
@JsonDeserialize(using = EnumerationDeserializer.class)
@EnumJson(serializeProjection = Projection.NAME, deserializationClass = Event.class)
public enum Event {
    FORGOT_PASSWORD("forgot password");

    private final String value;

    private Event(final String description) {
        this.value = description;
    }

}

这就是你要做的。

然后，如果您有一个“有”事件的类，您可以注释每个事件以您想要的方式序列化。

class EventHolder {
    @EnumJson(serializeProjection = Projection.NAME)
    Event someEvent;

    @EnumJson(serializeProjection = Projection.ORDINAL)
    Event someOtherEvent;

    @EnumJson(serializeProjection = Projection.VALUE)
    Event yetAnotherEvent;
}

我发现了一个非常好的和简洁的解决方案，特别是当您不能修改枚举类时，就像在我的例子中一样。然后，您应该提供一个启用了特定特性的自定义ObjectMapper。这些特性从Jackson 1.6开始就可以使用了。所以你只需要在你的enum中写入toString()方法。

public class CustomObjectMapper extends ObjectMapper {
    @PostConstruct
    public void customConfiguration() {
        // Uses Enum.toString() for serialization of an Enum
        this.enable(WRITE_ENUMS_USING_TO_STRING);
        // Uses Enum.toString() for deserialization of an Enum
        this.enable(READ_ENUMS_USING_TO_STRING);
    }
}

还有更多与枚举相关的特性可用，请参见这里:

https://github.com/FasterXML/jackson-databind/wiki/Serialization-Features https://github.com/FasterXML/jackson-databind/wiki/Deserialization-Features

可以采用多种方法将JSON对象反序列化为枚举。我最喜欢的风格是创建一个内部类:

import com.fasterxml.jackson.annotation.JsonCreator;
import com.fasterxml.jackson.annotation.JsonFormat;
import com.fasterxml.jackson.annotation.JsonProperty;
import org.hibernate.validator.constraints.NotEmpty;

import java.util.Arrays;
import java.util.Map;
import java.util.function.Function;
import java.util.stream.Collectors;

import static com.fasterxml.jackson.annotation.JsonFormat.Shape.OBJECT;

@JsonFormat(shape = OBJECT)
public enum FinancialAccountSubAccountType {
  MAIN("Main"),
  MAIN_DISCOUNT("Main Discount");

  private final static Map<String, FinancialAccountSubAccountType> ENUM_NAME_MAP;
  static {
    ENUM_NAME_MAP = Arrays.stream(FinancialAccountSubAccountType.values())
      .collect(Collectors.toMap(
        Enum::name,
        Function.identity()));
  }

  private final String displayName;

  FinancialAccountSubAccountType(String displayName) {
    this.displayName = displayName;
  }

  @JsonCreator
  public static FinancialAccountSubAccountType fromJson(Request request) {
    return ENUM_NAME_MAP.get(request.getCode());
  }

  @JsonProperty("name")
  public String getDisplayName() {
    return displayName;
  }

  private static class Request {
    @NotEmpty(message = "Financial account sub-account type code is required")
    private final String code;
    private final String displayName;

    @JsonCreator
    private Request(@JsonProperty("code") String code,
                    @JsonProperty("name") String displayName) {
      this.code = code;
      this.displayName = displayName;
    }

    public String getCode() {
      return code;
    }

    @JsonProperty("name")
    public String getDisplayName() {
      return displayName;
    }
  }
}

除了使用@JsonSerialize @JsonDeserialize，你还可以在对象映射器中使用SerializationFeature和DeserializationFeature (jackson绑定)。

例如DeserializationFeature。READ_UNKNOWN_ENUM_VALUES_USING_DEFAULT_VALUE，如果提供的枚举类型没有在枚举类中定义，则给出默认的enum类型。

@JsonFormat(shape = JsonFormat.Shape.OBJECT)
public enum LoginOptionType {

 PHONE(1, "Phone"), MAIL(2, "mail"), PERSONAL_EMAIL(3, "Personal email");

private static List<LoginOptionType> all;

static {
    all = new ArrayList<LoginOptionType>() {
        {
            add(LoginOptionType.PHONE);
            add(LoginOptionType.MAIL);
            add(LoginOptionType.PERSONAL_EMAIL);
        }
    };
}

private final Integer viewValue;

private final String name;

LoginOptionType(Integer viewValue, String name) {
    this.viewValue = viewValue;
    this.name = name;
}

public Integer getViewValue() {
    return viewValue;
}

public String getName() {
    return name;
}

public static List<LoginOptionType> getAll() {
    return all;
}
}

响应

[
{
    "viewValue": 1,
    "name": "Phone"
},
{
    "viewValue": 2,
    "name": "mail"
},
{
    "viewValue": 3,
    "name": "Personal email"
}
]