Question

我正在创建一个包含“活动”类别的应用，该数据使用JSON显示。在此，我使用Intent将ID字符串发送到另一个Activity 那么我只希望它显示包含cat_id = id（我发送到此活动）的数据

这是我的类别数组：-

[
  {
    "id": 1,
    "name": "Udemy Courses"
  },
  {
    "id": 2,
    "name": "Hindi movies"
  },
  {
    "id": 3,
    "name": "Hollywood Hindi"
  },
  {
    "id": 4,
    "name": "Hollywood English"
  },
  {
    "id": 5,
    "name": "Netflix Series"
  },
  {
    "id": 6,
    "name": "Tv Series"
  }
]

这是我的物品数组：-

[
  {
    "id": 1,
    "title": "first title",
    "description": "description description vdescription description description ",
    "date": "2019-01-08",
    "cat_id": 1
  },
  {
    "id": 2,
    "title": "second title",
    "description": "vdescription description vdescription description description ",
    "date": "2019-01-09",
    "cat_id": 2
  },
  {
    "id": 3,
    "title": "title",
    "description": "description description description description ",
    "date": "2019-01-15",
    "cat_id": 1
  },
  {
    "id": 4,
    "title": "title",
    "description": "description description description ",
    "date": "2019-01-09",
    "cat_id": 1
  }
]

我正在使用此代码

private int CategoryId;



CategoryId = Integer.parseInt(getIntent().getStringExtra("id"));





 @Override
        public void onResponse(JSONArray response) {
            JSONObject jsonObject = null;
            for (int i=0; i < response.length();i++){
                try {

                    jsonObject = response.getJSONObject(i);
if (jsonObject.getInt("cat_id")==CategoryId){
                        Item anime = new Item();

                        String Title = jsonObject.getString("title");
                        String Description = jsonObject.getString("description");
                        String Date = jsonObject.getString("date");

                        anime.setTitle(Title);
                        anime.setDescription(Description);
                        anime.setDate(Date);

                        listAnime.add(anime);
                        mProgress.dismiss();








                  } catch (JSONException e) {
                    e.printStackTrace();
                }
            }
            setuprecyclerview(listAnime);

        }
    }, new Response.ErrorListener() {
        @Override
        public void onErrorResponse(VolleyError error) {

        }
    });

    requestQueue = Volley.newRequestQueue(MainActivity.this);
    requestQueue.add(request);
}

private void setuprecyclerview(List<Category> listAnime) {
    RecyclerViewAdapter myAdapter = new RecyclerViewAdapter(this,listAnime);
    recyclerView.setLayoutManager(new LinearLayoutManager(this));
    recyclerView.setAdapter(myAdapter);
}

但这不会显示过滤的数据

例如：-当我单击id = 1（在类别数组中）的

然后它仅显示cat_id = 1的数据

但这显示了错误

Caused by: java.lang.NumberFormatException: Invalid int: "null"

Answer 1

我为您发表了评论

    @Override
    public void onResponse (JSONArray response){
        JSONObject jsonObject = null;
        for (int i = 0; i < response.length(); i++) {
            try {
                jsonObject = response.getJSONObject(i);

                //here you have all of the items and you should handle them

                if (jsonObject.getInt("cat_id") == 1) {

                    //change 1 to the number you want
                    //here you have the jsonObject with specific id
                    //for example, I want to get the description from jsonObject

                    String Desc = jsonObject.getString("description);
                }

                Category anime = new Category();

                anime.setName(jsonObject.getString("name"));

                listAnime.add(anime);
                mProgress.dismiss();
            } catch (JSONException e) {
                e.printStackTrace();
            }
        }
        setuprecyclerview(listAnime);

    }
},new Response.ErrorListener()

{
    @Override
    public void onErrorResponse (VolleyError error){

}
});

requestQueue =Volley.newRequestQueue(MainActivity .this);
requestQueue.add(request);
}

private void setuprecyclerview(List<Category> listAnime) {
    RecyclerViewAdapter myAdapter = new RecyclerViewAdapter(this, listAnime);
    recyclerView.setLayoutManager(new LinearLayoutManager(this));
    recyclerView.setAdapter(myAdapter);
}

如何从json数组获取自定义数据

1 个答案: