Deserialize JSON with Jackson without proper field

Question

I've got this JSON : {"success":false}

I want to deserialize this into this POJO :

class Message {
    private Map<String, String> dataset = new HashMap<String, String>();

    @JsonProperty("success")
    public boolean isSuccess() {
        return Boolean.valueOf(dataset.get("success"));
    }

    @JsonProperty("success")
    public void setSuccess(boolean success) {
        dataset.put("success", String.valueOf(success));
    }
}

Is it possible to deserialize this JSON into a class without field success? So far, i've always got the "UnrecognizedPropertyException: Unrecognized field "success""

Thanks for your help!

Answer 1

You could implement a method and annotate it with @JsonAnySetter like this:

@JsonAnySetter
public void handleUnknownProperties(String key, Object value) {
    // this will be invoked when property isn't known
}

another possibility would be turn this fail off like this:

objectMapper.configure(DeserializationConfig.Feature.FAIL_ON_UNKNOWN_PROPERTIES, false);

This would let you deserialize your JSON without failing when properties are not found.

Test

---

public static class Message {
    private final Map<String, String> dataset = new HashMap<String, String>();

    @Override
    public String toString() {
        return "Message [dataset=" + dataset + "]";
    }

}

@Test
public void testJackson() throws JsonParseException, JsonMappingException, IOException {
    String json = "{\"success\":false}";
    ObjectMapper om = new ObjectMapper().configure(DeserializationConfig.Feature.FAIL_ON_UNKNOWN_PROPERTIES, false);
    System.out.println(om.readValue(json, Message.class));
}