杰克逊 - 反序列化一个基础枚举

Question

是否可以反序列化具有一个索引的枚举？

enum Status {
  Active,
  Inactive
}

{status:1}表示Status.Active,但杰克逊使其成为Status.Inactive :(

Answer 1

public enum Status {
ACTIVE(1),
INACTIVE(2);
private final int value;
Status(int v) {
    value = v;
}
@org.codehaus.jackson.annotate.JsonValue
public int value() {
    return value;
}  
@org.codehaus.jackson.annotate.JsonCreator
public static Status fromValue(int typeCode) {
    for (Status c: Status.values()) {
        if (c.value==typeCode) {
            return c;
        }
    }
    throw new IllegalArgumentException("Invalid Status type code: " + typeCode);        

}}

Answer 2

您可以为枚举创建自定义类型反序列化器:

public enum Status {
    ACTIVE,
    INACTIVE;
    public static Status fromTypeCode(final int typeCode) {
        switch(typeCode) {
        case 1: return ACTIVE;
        case 2: return INACTIVE;
        }
        throw new IllegalArgumentException("Invalid Status type code: " + typeCode);
    }
}

public class StatusDeserializer extends JsonDeserializer<Status> {
    @Override
    public Status deserialize(final JsonParser parser, final DeserializationContext context) throws IOException {
        return Status.fromTypeCode(parser.getValueAsInt());
    }
}

然后,您可以告诉Jackson使用您的自定义deserialiser作为属性:

public class WarpDrive {
    private Status status; 
    @JsonDeserialize(using = StatusDeserializer.class)
    public void setStatus(final Status status) {
        this.status = status;
    }
    public Status getStatus() {
        return this.status;
    }
}

您还可以配置Jackson对象映射器,以便为所有出现的目标类型使用自定义反序列化器.请参阅http://wiki.fasterxml.com/JacksonHowToCustomDeserializers.