Android: converting JSON object values into specified class

Question

I am new to Android development and JAVA. I need to pass JSONObject values into android class. I am aware of the external libraries like GSON and others (like Jackson), but for this task I mus use only JSON. So far I have created an JSONObject:

public void jsonObject(){

    JSONObject test1 = new JSONObject();
    try{
        test1.put("name", "Pete");
        test1.put("surname", "Thompson");
        test1.put("age", "24");
        test1.put("height","182");
        test1.put("id", "45");
    }
    catch (JSONException a){
        a.printStackTrace();
    }
}

From this object, I need to pass the data to the following class members:

private String name;
private String surname;
private int age;
private int height;
private long id;

Thanks in advance.

Hey exactly this is provided by gson, have a look at some tuts out there, you will love it! — A.S.
– A.S., Commented Feb 26, 2014 at 13:13
Yeah I have used the GSON ext.lib., but now I have to use only built in lib. to better understand how it all works. — Ruslan
– Ruslan, Commented Feb 26, 2014 at 13:15

nikis · Accepted Answer · 2014-02-27 12:33:16Z

3

Then just make appropriate constructor for your class:

public class Test {
    String name;
    String surname;
    int age;
    int height;
    long id;

    public Test(JSONObject jsonObject) {
        this.name = jsonObject.getString("name");
        this.surname = jsonObject.getString("surname");
        this.age = jsonObject.getInt("age");
        this.height = jsonObject.getInt("height");
        this.id = jsonObject.getLong("id");
    }
}

edited Feb 27, 2014 at 12:33

answered Feb 26, 2014 at 13:15

nikis

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5 Comments

Ruslan Over a year ago

But the fields are private, the use of getters and setters is required , am I right?

nikis Over a year ago

@russell1911dev If you want to keep them private, yes, getters and setters are required (but please keep in mind, that they are not recommended in Android developer.android.com/training/articles/…). Or you can make them package-visible or public.

Ruslan Over a year ago

could you give me some example on how to put information from JSONObject, into getter and setter and print into console using system.out.print? I wouldn't ask, but I get nullpointerException.

nikis Over a year ago

@russell1911dev I've made them package-protected (with no access modifier), so you don't need getters and setters. Please note, that you cannot use System.out on Android, use Log class instead (developer.android.com/reference/android/util/Log.html)

nikis Over a year ago

If you get an error it's better to update your question.

Ahmed Zayed · Accepted Answer · 2014-02-26 13:15:45Z

1

try{    
    String name = test1.getString("name");
    String surName= test1.getString("surname");
    int age = test1.getInt("age");
    int height = test1.getInt("height");
    long id = test1.getLong("id");
    YourClass mInstance = new YourClass(name,surName,age,height,id);
}
catch(JSONParseException){
}

answered Feb 26, 2014 at 13:15

Ahmed Zayed

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Comments

A.S. · Accepted Answer · 2014-02-26 13:20:03Z

1

The workload when using GSON would look like this:

Object generated with GSON:

class MyUser{

    @Expose @SerializedName("name");
    private String name;

    @Expose @SerializedName("surname");
    private String surname;

    //and so on

    public String getName(){
        return name;
    }

    public static MyUser fromJson(String json){
        return new GsonBuilder().excludeFieldsWithoutExposeAnnotation()
                    .create().fromJson(json, MyUser.class);
    }

}

How to use:

class WorkWithObjects{
    private MyUser currentUser;

    public void dataRecieved(String json){
        currentUser = MyUser.fromJson(json);
        Log.i("NAME", "IS " + currentUser.getName());
    }

}

answered Feb 26, 2014 at 13:20

A.S.

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Comments

Dan Cuc · Accepted Answer · 2014-02-26 13:18:54Z

0

private String name;
private String surname;
private int age;
private int height;
private long id;

public JSONObject writeJsonObject(){

    JSONObject obj = new JSONObject();
    try{
        obj.put("name", name);
        obj.put("surname", surname);
        obj.put("age", age);
        obj.put("height",height);
        obj.put("id", id);
    }
    catch (JSONException a){
        a.printStackTrace();
    }
    return obj ;
}
public Test(JSONObject obj) {
    this.name = obj.getString("name");
    this.surname = obj.getString("surname");
    this.age = obj.getInt("age");
    this.height = obj.getInt("height");
    this.id = obj.getLong("id");
}

answered Feb 26, 2014 at 13:18

Dan Cuc

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