Convert JSON object to simple variables in Java

Question

I have a very heavy JSON with lots of parameters which I want to convert to a Java object containing just a few of them.

Direct conversion such as this

DataObject obj = gson.fromJson(br, DataObject.class);

is not an option.

How can I access individual fields within objects (just value under date and type under attributes under completion_date)? JSON example:

{"results":[
    {"date":{
        "attributes":{
        "type":null},
        "value":"December 13, 2010"},
    "is_structured":"No",
    "completion_date":{
        "attributes":{
            "type":"Anticipated"},
        "value":"March 2016"},
     ....

May be you are talking about JSON.parse look at this stackoverflow.com/questions/4935632/parse-json-in-javascript — Hisham
– Hisham, Commented Apr 1, 2015 at 13:35
So you have your json as String and you want to access given property directly? Or you have some sort of Map or JsonObject? — zlobi.wan.kenobi
– zlobi.wan.kenobi, Commented Apr 1, 2015 at 13:42

ceekay · Accepted Answer · 2015-04-02 10:17:24Z

4

If you don't want to directly convert your input to the expected Object you could create a JSONObject from your input like

JSONObject root = new JSONObject(input);

and then navigate from there exactly to the attribute you need e.g.:

root.getJSONArray("results").getJSONObject(0).getString("is_structured");

EDIT: (receive value under date)

root.getJSONArray("results").getJSONObject(0).getJSONObject("date").getString("value");

edited Apr 2, 2015 at 10:17

answered Apr 1, 2015 at 13:41

ceekay

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3 Comments

FloIancu Over a year ago

Thank you for the answer! And what is the syntax for getting a sub-field, for example date.value?

ceekay Over a year ago

Therefore you have to navigate into the deeper JSONObject. See my edit

FloIancu Over a year ago

Worked perfectly. Thank you for taking your time to explain it to me, I definitely grasp the matter now!

Raffaele · Accepted Answer · 2015-04-01 13:41:30Z

0

You can have just the fields you need in your bean definition. For example, given the following JSON

{
  firstName: 'It',
  lastName: 'works',
  ignoreMe: '!!!'
}

the bean definition does not contain ignoreMe:

public class Contact {
    public String firstName; // setter and getter
    public String lastName;  // setter and getter
}

But it works with Gson:

Gson gson = new Gson();
Contact contact = gson.fromJson(new FileReader(new File("test.json")), Contact.class);
System.out.println("It worked!");

answered Apr 1, 2015 at 13:41

Raffaele

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