Java Spring Boot - api对象创建

Question

我正在开发 Java Spring boot api。

当调用获取/api/home 时

我想返回这个 json 示例结构。

      var response = [
            {
              "type": "profile-breakdown",
              "order": 0,
              "grid-width": 6,
              "grid-background": "",
              "grid-background-url": "",
              "title": "",
              "contents": {
                "name": "Name1",
                "avatar" : 1,
                "nextSDQ": 4,
                "SQDCount": 3
              }
            },
            {
              "type": "current-standing",
              "order": 1,
              "grid-width": 6,
              "grid-background": "",
              "grid-background-url": "",
              "title": "Your current standing summary",
              "contents": {
                "0": ["emotional distress", "behavioural difficulties", "hyperactivity and concentration difficulties", "difficulties in getting along with other young people"],
                "4": ["kind and helpful behaviour"]
              }
            }
]

--我一直在构建各种函数来获取“配置文件分割”和“当前状态”——我想将响应附加到这些函数以模仿上述结构。

因此，在 MyService 中/api/home 获取 RequestMapped 时，我开始 Hook 到我的类 MyApiHome

    MyApiHome myApiHome = new MyApiHome();
    JSONObject homeObj = myApiHome.getHomeData();

在 MyApiHome 中 - 我想将 getHomeData 中的“homeObj”设为数组，而不是 JSONOBject - 但随后我开始陷入强制类型转换等麻烦。我想以这样的方式构建它 - 如果 getProfileBreakDown为空或解耦，它不会附加到 homeObj。

public class MyApiHome {

    @SuppressWarnings("unchecked")
    public JSONObject getHomeData(){
        //build clean home object
        JSONObject homeObj = new JSONObject();              
        homeObj.put("profile", this.getProfileBreakDown());
        homeObj.put("currentstanding", this.getCurrentStanding());
        //HashMap<List<String>, Object> hashMap = new HashMap<List<String>, Object>();
                //hashMap.put())


        return homeObj;
    }

    @SuppressWarnings("unchecked")
    public Object getProfileBreakDown(){
        //build clean home object
        JSONObject contents = new JSONObject();     
        contents.put("name", "Name1");
        contents.put("avatar", 1);
        contents.put("nextSDQ", 4);
        contents.put("SQDCount", 3);

        //build clean home object
        JSONObject json = new JSONObject();             
        json.put("type", "profile-breakdown");
        json.put("order", 0);
        json.put("grid-width", 6);
        json.put("grid-background", "");
        json.put("grid-background-url", "");
        json.put("title", "");
        json.put("contents", contents);

        return json;
    }


    @SuppressWarnings("unchecked")
    public Object getCurrentStanding(){

        String[] stressArray1 = {"emotional distress", "behavioural difficulties", "hyperactivity and concentration difficulties", "difficulties in getting along with other young people"};
        String[] stressArray2 = {"kind and helpful behaviour"};


        //build clean home object
        JSONObject contents = new JSONObject();     
        contents.put("0", stressArray1);
        contents.put("4", stressArray2);

        //build clean home object
        JSONObject json = new JSONObject();             
        json.put("type", "current-standing");
        json.put("order", 1);
        json.put("grid-width", 6);
        json.put("grid-background", "");
        json.put("grid-background-url", "");
        json.put("title", "Your current standing summary");
        json.put("contents", contents);

        return json;
    }

}

Answer 1

要创建 JSON 数组，我们需要使用包含 JSONObject 列表的 JSONArray 对象。