Question

在我的代码中，我试图从MySQL数据库中填充值微调器，但是我无法获得该值。

我的jsonArray出了什么问题？

java.lang.String无法转换为JSONObject

JsonArrayRequest jsonSpinnerObjectRequest = new JsonArrayRequest(fetch_city_url, new Response.Listener<JSONArray>() {
        @Override
        public void onResponse(JSONArray response) {
            try {
                JSONObject jsonObject = null;
                JSONArray jsonArray = jsonObject.getJSONArray("result");
                for (int i = 0; i < jsonArray.length(); i++) {

                    JSONObject city = jsonArray.getJSONObject(i);
                    Log.i("onResponseSpinner", city.getString("kd_kelas") + " " + city.getString("kelas"));
                    kelasList.add(city.getString("kelas"));
                }
                spinnerAdapter.notifyDataSetChanged();

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

        }
    }, new Response.ErrorListener() {
        @Override
        public void onErrorResponse(VolleyError error) {
            Log.e("VOLLEY","ERROR in response"+error.getMessage());
        }
    });

Answer 1

在这里，您将JSON对象设置为null，然后尝试从中获取JSON数组：

JSONObject jsonObject = null;
JSONArray jsonArray = jsonObject.getJSONArray("result");

我认为你想要这样的东西：

public void onResponse(JSONObject response) {
    try {
        JSONObject jsonObject = new JSONObject(response);
        JSONArray = jsonArray = new jsonObject.getJSONArray("result");
        for (int i = 0; i < jsonArray.length(); i++) {
            JSONObject city = jsonArray.getJSONObject(i);
            Log.i("onResponseSpinner", city.getString("kd_kelas") + " " + city.getString("kelas"));
            kelasList.add(city.getString("kelas"));
        }
        spinnerAdapter.notifyDataSetChanged();
        } catch (JSONException e) {
            e.printStackTrace();
        }

Answer 2

查看您的代码，我认为您已经在响应中获取JSON数组，因此只需从此JSON数组中获取值，请查看以下代码

JsonArrayRequest jsonSpinnerObjectRequest = new JsonArrayRequest(fetch_city_url, new Response.Listener<JSONArray>() {
    @Override
    public void onResponse(JSONArray response) {
        try {

            for (int i = 0; i < response.length(); i++) {

                JSONObject city = response.getJSONObject(i);
                Log.i("onResponseSpinner", city.getString("kd_kelas") + " " + city.getString("kelas"));
                kelasList.add(city.getString("kelas"));
            }
            spinnerAdapter.notifyDataSetChanged();

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

    }
}, new Response.ErrorListener() {
    @Override
    public void onErrorResponse(VolleyError error) {
        Log.e("VOLLEY","ERROR in response"+error.getMessage());
    }
});

我得到关于java.lang.String的错误无法转换为JSONarray

2 个答案: