是否有可能反序列化枚举,它有一个基于1的索引?
enum Status {
Active,
Inactive
}
{ Status.Active :1}的意思是状态,但是Jackson把它变成了Status.Inactive :(
发布于 2012-02-13 18:53:46
您可以为枚举创建自定义类型反序列化程序:
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对属性使用自定义的反序列化程序:
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;
}
}
您还可以将Jackson对象映射器配置为对所有出现的目标类型使用您的自定义反序列化程序。参见http://wiki.fasterxml.com/JacksonHowToCustomDeserializers。
发布于 2012-10-11 22:58:10
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);
}}
https://stackoverflow.com/questions/8790389
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