Deserialize wrapped list using Jackson

Question

I have a JSON Object like this:

{"geonames":[
   {"countryId":"2017370",
    "adminCode1":"73"},
   {"countryId":"2027370",
    "adminCode1":"71"},
    ...]}

How can i deserialize this object DIRECTLY to List<GeoName>, ignoring the first layer (geonames wrapper), instead of deserializing to a wrapper object containing List<GeoName> as @JsonProperty("geonames")?

if you are getting this JSON from any rest endpoint then try spring rest template, it has messgae converters which will take care of marshalling and unmarshalling — ppuskar, Dec 30 '14 at 04:45

score 4 · Accepted Answer · answered Dec 30 '14 at 04:25

4

Use an ObjectReader with a root name

ObjectMapper mapper = new ObjectMapper();
ObjectReader reader = mapper.reader(new TypeReference<List<GeoName>>() {}).withRootName("geonames");
List<GeoName> list = reader.readValue(json);

answered Dec 30 '14 at 04:25

Sotirios Delimanolis

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Is there any way to achieve this with annnotations? Without instantiating a reader? – localhost Dec 30 '14 at 07:18
@localhost Not while deserializing to `List` directly. If you could change the root name to `List`, you could use `mapper.configure(DeserializationFeature.UNWRAP_ROOT_VALUE, true);` and use the mapper directly. – Sotirios Delimanolis Dec 31 '14 at 00:59
@localhost But what's wrong with `ObjectReader`? These are generated internally anyway. – Sotirios Delimanolis Dec 31 '14 at 00:59
can't change the root name. Thx for answer, i accept it :) – localhost Dec 31 '14 at 02:13

Deserialize wrapped list using Jackson

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