Deserializing JSON property as String with Jackson

A while ago I faced a situation that it was necessary to parse a JSON like this:

{
  "payload": {
    "foo": "bar",
    "biz": [
      "lorem",
      "ipsum"
    ]
  },
  "signature": "somehmacsign"
}

Into this Java class:

public class Request {

    private String payload;
    private String signature;

    // Getters and setters omitted
}

The challenge here is to read the content of the payload property (which can be any valid JSON object/array/value) into a string. That is, the content of the payload property must not be parsed into a POJO.

It could be achieved with the following custom deserializer:

public class RawJsonDeserializer extends JsonDeserializer<String> {

    @Override
    public String deserialize(JsonParser jp, DeserializationContext ctxt)
           throws IOException, JsonProcessingException {

        ObjectMapper mapper = (ObjectMapper) jp.getCodec();
        JsonNode node = mapper.readTree(jp);
        return mapper.writeValueAsString(node);
    }
}

Then annotate the payload attribute with the @JsonDeserialize annotation, referencing the deserializer created above:

public class Request {

    @JsonDeserialize(using = RawJsonDeserializer.class)
    private String payload;

    private String signature;

    // Getters and setters omitted
}

And parse the JSON using ObjectMapper:

String json = "{\"payload\":{\"foo\":\"bar\",\"biz\":[\"lorem\","
            + "\"ipsum\"]},\"signature\":\"somehmacsign\"}";

ObjectMapper mapper = new ObjectMapper();
Request request = mapper.readValue(json, Request.class);

Since the deserializer defined above will parse the JSON into the Jackson tree model and then serialize it back to a string, the result may not be equal to the original content of the payload property. If you want to keep the spaces, tabs, line breaks and the order of the JSON properties, have a look at the following implementation, which won’t parse the JSON into the Jackson tree model:

public class RawJsonDeserializer extends JsonDeserializer<String> {

    @Override
    public String deserialize(JsonParser jp, DeserializationContext ctxt)
           throws IOException, JsonProcessingException {

        long begin = jp.getCurrentLocation().getCharOffset();
        jp.skipChildren();
        long end = jp.getCurrentLocation().getCharOffset();

        String json = jp.getCurrentLocation().getSourceRef().toString();
        return json.substring((int) begin - 1, (int) end);
    }
}

In both deserializers, an exception will be thrown if the JSON is invalid.