How to form JSON in Java with multi rows in the lower level

Question

I'm new in Java and JSON and I'm trying to form JSON for my rest api (I previously used XML and converter cause I'm using AngularJS, but I would like to form JSON without conversioning). The format of the JSON should be:

{
  "friends": {
    "friend": [
      {
        "email": "aa@aa.aa",
        "firstName": "aa",
        "id": "31",
        "lastName": "aa"
      },
      {
        "email": "bb@bb.bb",
        "firstName": "bb",
        "id": "32",
        "lastName": "bb"
      }
    ]
  }
}

The problem is (I googled a lot, and the best article is - Java & Json: Wrapping elements when serialising) but in the example - I got only one row in the lower level and if I'm using something like this:

    public class JsonSimpleExample {
     public static void main(String[] args) {
         org.json.simple.JSONObject jsonObject3=new org.json.simple.JSONObject();
         org.json.simple.JSONObject jsonObject2=new org.json.simple.JSONObject();
         for (int j = 0; j < 4; j++) {
         org.json.simple.JSONObject jsonObject=new org.json.simple.JSONObject();

         Map<String, Object> data = new HashMap<String, Object>();
            data.put("email","aa@aa.aa");
            data.put("firstName","aa");
            data.put("lastName","aa");
            data.put("id",j);
            JSONObject json = new JSONObject();
            json.putAll( data );
         jsonObject2.put("friend",json); 
         jsonObject3.put("friends",jsonObject2); 
    }

    System.out.print(jsonObject3);

}

Here I got only the last row, what is obvious. For the other hand - while I'm using HashMap or Map - I got a little bit different format. May be I should use something from this example (I mean List), but not sure how it finally should be: https://stackoverflow.com/questions/27743227/creating-parent-property-in-json. So, I implemented it this way:

public class JsonSimpleExample {
     public static void main(String[] args) {
         org.json.simple.JSONObject jsonObject3=new org.json.simple.JSONObject();
         org.json.simple.JSONObject jsonObject2=new org.json.simple.JSONObject();

         JSONArray arr = new JSONArray();
         HashMap<String, JSONObject> map = new HashMap<String, JSONObject>();
            for(int i = 0 ; i < 10 ; i++) {
                JSONObject data=new JSONObject();
                data.put("email","aa@aa.aa"+i);
                data.put("firstName","aa");
                data.put("lastName","aa");
                data.put("id",i);
                map.put("json" + i, data);
                arr.put(map.get("json" + i));
              }

         jsonObject2.put("friend",arr); 
         jsonObject3.put("friends",jsonObject2); 

        System.out.print(jsonObject3);
    }
    }

I don't care for right now about how to generate JSON (I mean - using Jackson, Pojo or something else), my main problem is to understand - if I'm on the right way on getting the same data format as I had after XML to JSON data conversion. And what could I improve in my code?

Thank you!

score 0 · Answer 1 · answered May 11 '16 at 01:18

0

Create a Friend class which has fields for email, id, firstName, lastName, etc. Populate a collection object with Friend class instances. Serialize the collection.

answered May 11 '16 at 01:18

NicoE

186
7

score 0 · Accepted Answer · answered May 11 '16 at 04:00

0

Serializing and Deserializing JSON objects are quite easy when you use some existing libraries (why to re-invent wheel?). What you could improve is to use gson - Google's library for working with json. I will explain how it works:-

Create a Friend Class with all the fields you require.

Class Friend{
  String email;
  String firstName;
  Integer id;
  String lastName;

  Friend(String email,String firstName,Integer id,String lastName){
    this.email = email;
    this.firstName = firstName;
    this.id = id;
    this.lastName = lastName;
  //This is Constructor
  }
}

Create a class for you enclosing JSON Object friends:

Class Friends{
  Friend[] friend;

  Friends(Friend[] friend){
     this.friend = friend;
  }
}

Now say obj is of type Friends. You will simple get the JSON string by:

import com.google.gson.Gson;
Gson gson = new Gson();
String jsonInString = gson.toJson(obj);

References: http://www.mkyong.com/java/how-do-convert-java-object-to-from-json-format-gson-api/

Let me know if you still have trouble making it work!

answered May 11 '16 at 04:00

Mangat Rai Modi

5,397
8
45
75

Mangat, I implemented 2 separate classes as you described. But the problem is - I got code: `import com.google.gson.*; public class Gson { public static void main(String[] args) { Friend friend = new Friend(); friend.setEmail("email"); friend.setFirstName("first"); friend.setLastName("last"); friend.setId(2); Gson gson = new Gson(); String json = gson.toJson(friend); } }` but I got an error here: at *toJson* I got an error - The method toJson(Friend) is undefined for the type Gson. I'm using gson-2.6.2.jar library. Thank you! – Aleksey Kiselev May 13 '16 at 00:13
You don't need to create the Gson jar. Its already imported by import com.google.gson.Gson; When you created Gson class, it has masked the original Gson class from google – Mangat Rai Modi May 13 '16 at 03:54
Oh, I got it (I made it by mistake, just decided that this gonna be a good name for my class).. Thank you, Mangat – Aleksey Kiselev May 13 '16 at 17:19
Just one last thing - how could I add top level "Friend" tag in this code: `public class jsonnew { public static void main(String[] args) { //Friends friends = new Friends(null); List friends = new ArrayList(); for (int i = 0; i < 3; i++) { Friend friend = new Friend(); friend.setEmail("email"); friend.setFirstName("first"); friend.setLastName("last"); friend.setId(i); friends.add(friend); } Gson gson = new Gson(); String json = gson.toJson(friends); System.out.println(json); } }` Thank you! – Aleksey Kiselev May 14 '16 at 01:01

How to form JSON in Java with multi rows in the lower level

2 Answers2