Java: Append JSON Object to a JSON Object

Question

I want to store chat history in a JSON Object, one object for each conversation, within this object i want to have an Array with all the messages.

Like this:

{"Channel_123":[
           {"from":"john","to":"bill", "msg":"Hello", "time":"09:57"},
           {"from":"bill","to":"john", "msg":"Hey John", "time":"09:58"}
         ]
}, {"Channel_234":[
           {"from":"bob","to":"judy", "msg":"Hello", "time":"10:37"},
           {"from":"judy","to":"bob", "msg":"Hey!", "time":"10:38"}
         ]
}

My current method looks like this: (the string channel contains the conversation id illustrated above with Channel_234

JSONObject obj = new JSONObject();
JSONObject msgObj = new JSONObject();

public void addMessage(String from, String to, String msg, String time, String channel) {

    try {
        msgObj.put("from", from);
        msgObj.put("to", to);
        msgObj.put("msg", msg);
        msgObj.put("time", time);


        obj.put(channel, array);

    } catch (Exception ex) {
        System.out.println(ex.getStackTrace());
        logger.log(Level.SEVERE, "Exception: ", ex);
    }

    System.out.println(obj);
}

But for some reason the method is not appending to the object it is overwriting what was previously there.

Any ideas?

EDIT

I am using the simple json lib: json-simple-1.1.1.jar

did you checkout whether the channel variable doesn't have same value??? — sam
– sam, Commented Dec 3, 2013 at 9:05
It has a value but its being overwritten each time there is a new message, so always its just one message being saved — Alosyius
– Alosyius, Commented Dec 3, 2013 at 9:06
you have problem in create an object, each time object is recreated then only you get current value. — Mr.G
– Mr.G, Commented Dec 3, 2013 at 9:09
your key is channel variable if the channel variable has same values then they will be overwritten.. since you are not appending the data — sam
– sam, Commented Dec 3, 2013 at 9:10
Changing obj.put to obj.append seems to have solved the problem! — Alosyius
– Alosyius, Commented Dec 3, 2013 at 9:12

Damian Leszczyński - Vash · Accepted Answer · 2013-12-03 09:28:28Z

1

I am not familiar with the jason-simple API but you should create a new instance of JSONObject for each item.

    private JSONObject createMesssage(String form, String to, String msg, String time) throws Exception {

       JSONObject jsonMessage= new JSONObject();

            jsonMessage.put("from", from);
            jsonMessage.put("to", to);
            jsonMessage.put("msg", msg);
            jsonMessage.put("time", time);

       return jsonMessage;  

    }

  JSONObject obj = new JSONObject();

  private void addMessage(String channel, JSonObject messsage) throw Exception {
               obj.put(channel, message);
   }

  public void saveMessage(String form, String to, String msg, String time, Striing channel ) {

    try {
      addMessage(chanell,createMessage(form,to,msg,time));

    } catch (Exception ex) {
        logger.log(Level.SEVERE, "Exception: ", ex);
    }
  }

edited Dec 3, 2013 at 9:28

answered Dec 3, 2013 at 9:20

Damian Leszczyński - Vash

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Comments

Boris Brodski · Accepted Answer · 2013-12-06 16:00:18Z

0

A valid JSON data representation won't duplicate keys in the same nesting level. Keeping in mind that only numbers, strings, null, objects and arrays are valid JSON primitives, you have some structure like:

[
  {
    ...
    "id_str" : "foo",
    ...
  },
  {
    ...
    "id_str" : "bar",
    ...
  },
  ...
]

Which is an array of JSON objects. In this case, you'd have to index the 4th element of the array and then index the "id_str" attribute of that object. Check the documentation of whatever JSON library you're using to decode the data to figure out how to do that.

edited Dec 6, 2013 at 16:00

Boris Brodski

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answered Dec 3, 2013 at 9:15

user3060664

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