How to deserialize a JSON to an object ignoring a key using Jackson?

Question

Lets say I have a JSON:

{

"MovieCount": 153,
"MoviesList": [{...},{...},...]
}

I got:

mapper.readValue(moviesListArrayString, new TypeReference<List<Movie>>(){});

But that works when moviesListArrayString is:

[{...},{...},...]

if my String is the original JSON, how can I tell Jackson to ignore MovieCount and deserialize MoviesList?

Answer 1

One solution can be to read the full JSON as a JsonNode and only deserialize what you are interested in:

final JsonNode node = mapper.readTree(...);
movieList = mapper.readValue(node.get("MoviesList").traverse(), typeRefHere);

Answer 2

I would create a wrapper object with two properties:

• Number movieCount
• List<Movie> moviesList

Then read the value as YourWrapperObject.class and do whatever you need to do with the moviesList property value, while ignoring the moviecount property.

Quick, very ugly but functional example

package test;

import java.util.List;

import com.fasterxml.jackson.annotation.JsonProperty;
import com.fasterxml.jackson.databind.ObjectMapper;

public class Main {

    public static void main(String[] args) {
        ObjectMapper mapper = new ObjectMapper();

        String json = "{\"MovieCount\": 153,\"MoviesList\": [{},{}]}";
        try {
            List<Movie> movies = ((MovieWrapper) mapper.readValue(json, MovieWrapper.class)).moviesList;
            System.out.println(movies.size());
        }
        catch (Throwable t) {
            t.printStackTrace();
        }

    }

    static class MovieWrapper {

        @JsonProperty(value = "MovieCount")
        int movieCount;
        @JsonProperty(value = "MoviesList")
        List<Movie> moviesList;
    }

    static class Movie {

    }

}

Output

2