Java:将 JSON 对象附加到 JSON 对象

Question

我想将聊天历史记录存储在一个 JSON 对象中，每个对话一个对象，在这个对象中我想有一个包含所有消息的数组。

像这样:

{"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"}
         ]
}

我当前的方法如下所示:(字符串 channel 包含上面所示的对话 ID 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);
}

但由于某种原因，该方法没有附加到对象，而是覆盖了之前的内容。

有什么想法吗？

编辑

我正在使用简单的 json 库:json-simple-1.1.1.jar

Answer 1

我不熟悉 jason-simple API，但您应该为每个项目创建一个新的 JSONObject 实例。

    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);
    }
  }