On Mon, Sep 10, 2018 at 8:13 AM Slawek Mazur <[email protected]> wrote:
>
> Here's the scenario:
>
> public class UnwrappedWithPropertyName {
>
> public static void main(String[] args) throws JsonProcessingException {
>
> final Address address = new Address(new Postcode("45678"));
>
> final ObjectMapper mapper = new ObjectMapper();
>
> System.out.println(mapper.writeValueAsString(address));
> }
>
>
> static class Address {
>
> @JsonUnwrapped
> @JsonProperty("postcode")
> private final Postcode postcode;
>
> Address(Postcode postcode) {
> this.postcode = postcode;
> }
>
> public Postcode getPostcode() {
> return postcode;
> }
> }
>
> static class Postcode {
>
> private final String value;
>
> Postcode(String value) {
> this.value = value;
> }
>
> public String getValue() {
> return value;
> }
> }
> }
>
> That results in: {"value":"45678"} what I would expect is to see it as
> {"postcode":"45678"}
`@JsonUnwrapped` "peels off" surrounding property, and then all
properties within `Postcode` are
serialized with settings they have. So `@JsonProperty` is ignored, and
that is as designed from
Jackson perspective.
But to achieve result you want, you would be better off using
`@JsonValue` on `Postcode`:
static class Postcode {
private final String value;
// optional: to deserialize from String value
@JsonCreator(mode = JsonCreator.Mode.DELEGATING)
Postcode(String value) {
this.value = value;
}
@JsonValue
public String getValue() {
return value;
}
}
in which `Postcode` is serialized as String (and deserialized from
one); you can then still rename
(or not) property that points to it as you wish.
I hope this helps,
-+ Tatu +-
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