CODEWITHSUNDEEP
javajson
FloIancu
FloIancu

Reputation: 2725

Convert JSON object to simple variables in Java

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"},
     ....

Upvotes: 2

Views: 5786

Answers (2)

ceekay
ceekay

Reputation: 466

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");

Upvotes: 4

Raffaele
Raffaele

Reputation: 20885

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!");

Upvotes: 0

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