is there any simple example for Android of using JSON in a serialization?
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4 Answers
We use the gson library for that. Serialization is as simple as calling
new Gson().toJson(obj)
And for deserialization,
new Gson().fromJson(jsonStr, MyClass.class);
1 Comment
Chris Ho
deserialization : MyClass myClass = new Gson().fromJson(jsonStr, MyClass.class);
If you want to avoid using another library in your Android project just to (de)serialize JSON, you cau use following code as I do.
To serialize
JSONObject json = new JSONObject();
json.put("key", "value");
// ...
// "serialize"
Bundle bundle = new Bundle();
bundle.putString("json", json.toString());
and to deserialize
Bundle bundle = getBundleFromIntentOrWhaterver();
JSONObject json = null;
try {
json = new JSONObject(bundle.getString("json"));
String key = json.getString("key");
} catch (JSONException e) {
e.printStackTrace();
}
2 Comments
Pure Function
This is cool, but I can't imagine it working well for larger sets of complicated JSON
Martin Edlman
You're absolutely right, this is suitable for quite simple and reasonably small JSONs. :-)
There is a simple library to (de)serialize JSON,compatible with android own json library.
// deserialize a java bean to json object
JSONObject studentJson = JsonDeer.toJson(student);
// serialize a java bean from json object
Student student1 = JsonDeer.fromJson(studentJson,Student.class);
Comments
protected void onPostExecute(String results) {
if (results!=null) {
try {
Tec tec_m=new Tec();
tec_m=new Gson().fromJson(results, Technician.class);
((AndroidActivity)activity).setData(tec_m);
} catch (JSONException e) {
// TODO Auto-generated catch block
e.printStackTrace();
}
