我有这么一个json字符串通过fastjson的parse方法接的 我有什么办法遍历他们

我有这么一个json字符串通过fastjson的parse方法接的
我有什么办法遍历他们并且我要用对应的每个(edyear-bgyear)*money
然后累加

{"bgyear1":"2019","edyear2":"2019","edyear3":"2019","edyear1":"2019","Money3":"33333","Money1":"11111","Money2":"2222","bgyear2":"2019","bgyear3":"2019"}

如上图，代码如下

//1、定义json数据
        String json="{\"bgyear1\":\"2019\",\"edyear2\":\"2019\",\"edyear3\":\"2019\",\"edyear1\":\"2019\",\"Money3\":\"33333\",\"Money1\":\"11111\",\"Money2\":\"2222\",\"bgyear2\":\"2019\",\"bgyear3\":\"2019\"}";
        //2、解析
        JSONObject jsonObject= JSONObject.parseObject(json);
        Map<String,Integer> resMap=new HashMap<>();
        //3、从1开始,依次循环取出json中字段
        for(int i=1;;i++){
            Integer edyear = jsonObject.getInteger("edyear"+i);
            Integer bgyear = jsonObject.getInteger("bgyear"+i);
            Integer money = jsonObject.getInteger("Money"+i);
            //3.1、如果任一字段为null,则打断循环
            if(edyear==null||bgyear==null||money==null){
                break;
            }
            //3.2、添加结果
            resMap.put("res"+i,(edyear-bgyear)*money);
        }
        //4、打印结果(注意此处使用的是jdk1.8语法lambda表达式)
        resMap.forEach((k,v)->{
            System.out.println(k+":"+v);
        });

String jsons = "{\"bgyear1\":\"2019\",\"edyear2\":\"2019\",\"edyear3\":\"2019\",\"edyear1\":\"2019\",\"Money3\":\"33333\",\"Money1\":\"11111\",\"Money2\":\"2222\",\"bgyear2\":\"2019\",\"bgyear3\":\"2019\"}";
        JSONObject jsonObject = JSON.parseObject(jsons);
        jsonObject.entrySet().forEach(V->{
            System.out.println(V);
        });

String str = "{\"bgyear1\":\"2019\",\"edyear2\":\"2019\",\"edyear3\":\"2019\",\"edyear1\":\"2019\",\"Money3\":\"33333\",\"Money1\":\"11111\",\"Money2\":\"2222\",\"bgyear2\":\"2019\",\"bgyear3\":\"2019\"}";
JSONObject obj = JSON.parseObject(str);
int edyear1 = Integer.parseInt(obj.get("edyear1").toString());
int bgyear1 = Integer.parseInt(obj.get("bgyear1").toString());
int money1 = Integer.parseInt(obj.get("Money1").toString());
int result = (edyear1 - bgyear1) * money1;