How to deserialize a JSON HashMap?

Question

This is my situation:

private Map<String, Object> data = new HashMap<>();

public void setName(String x) {put("name", x);}
public String getName() {return (String)get("name");}

public void setAge(Integer x) {put("age", x);}
public Integer getAge() {return (Integer)get("age");}

public void setUUID(UUID x) {put("uuid", x);}
public UUID getUUID() {return (UUID)get("uuid");}

Now when I generate JSON from this object I get {"data":{"name": ...}} I don't want the "data":{} part.

How to fix this?

[EDIT] When this class is exported as JSON by Spring the format is exactly what I want.

how are you generating the JSON object? Normally there is no data part. — Jason, Apr 15 '16 at 17:32
There are a few solutions [HERE](http://stackoverflow.com/questions/4802887/gson-how-to-exclude-specific-fields-from-serialization-without-annotations). You can mark it as transient, or you can expose the ones you wish and then grab them only via your Gson API. — ManoDestra, Apr 15 '16 at 17:56
@Expose has no effect. When I use transient my JSON is empty {}. — Ricardo, Apr 15 '16 at 19:44

score 0 · Answer 1 · answered Apr 15 '16 at 17:48

Since your entire class seems to be inside a map, simply add a getter method for the Map to you class and instead of serializing the class itself, do it like:

gson.toJson(foo.getData())

On a side note, I gotta say it is pretty awkward having getters and setters doing actions on a map.

score 0 · Accepted Answer · answered Apr 15 '16 at 19:58

0

This works for me.

    ObjectMapper mapper = new ObjectMapper();
    String json = mapper.writeValueAsString(myClass);

answered Apr 15 '16 at 19:58

Ricardo

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How to deserialize a JSON HashMap?

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