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Jackson - deserialize one base enums

is it possible to deserialize enums, which have a one based index?

enum Status {
  Active,
  Inactive
}

{status:1} means Status.Active, but Jackson makes it Status.Inactive :(

like image 204
Dirk Avatar asked Jan 09 '12 15:01

Dirk


2 Answers

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);        

}}
like image 104
user1738531 Avatar answered Oct 18 '22 04:10

user1738531


You can create a custom type deserialiser for your enum:

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());
    }
}

You can then tell Jackson to use your custom deserialiser for a property:

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;
    }
}

You can also configure the Jackson object mapper to use your custom deserialiser for all occurrences of the target type. See http://wiki.fasterxml.com/JacksonHowToCustomDeserializers.

like image 35
Nathan Avatar answered Oct 18 '22 06:10

Nathan