Question

我正在尝试从如下所示的JSON数组中获取JSON数据：

{
"common": [
    {
        "food_name": "eggs",
        "serving_unit": "large",
        "tag_name": "raw eggs",
        "serving_qty": 1,
        "common_type": null,
        "tag_id": "775",
        "photo": {
            "thumb": "https://d2xdmhkmkbyw75.cloudfront.net/775_thumb.jpg"
        },
        "locale": "en_US"
    },

这就是我正在使用的：

    public class GetDietData extends AsyncTask<String, Void, String> {

    @Override
    protected String doInBackground(String... strings) {

        String calories = "UNDEFINED";
        try {

            URL urlForGetRequest = new URL("https://trackapi.nutritionix.com/v2/search/instant?query=egg");
            HttpURLConnection connection = (HttpURLConnection) urlForGetRequest.openConnection();
            connection.setRequestMethod("GET");
            connection.setRequestProperty("x-app-key", "REMOVED");
            connection.addRequestProperty("x-app-id", "REMOVED");

            InputStream stream = new BufferedInputStream(connection.getInputStream());
            BufferedReader bufferedReader = new BufferedReader(new InputStreamReader(stream));
            StringBuilder builder = new StringBuilder();

            String inputString;
            while ((inputString = bufferedReader.readLine()) != null) {
                builder.append(inputString);
            }


            JSONObject jsonRes = new JSONObject();
            JSONArray common = jsonRes.getJSONArray("common");

            for (int i=0; i<common.length(); i++)
            {
                JSONObject jsonObj = common.getJSONObject(i);

                 calories = jsonObj.getString("food_name");

            }

            connection.disconnect();
        } catch (IOException | JSONException e) {
            editText=(findViewById(R.id.editTextDiet));
            e.printStackTrace();
        }


        return calories;

    }


    @Override
    protected void onPostExecute(String calories) {

        if (calories == "UNDEFINED") {
            Toast.makeText(Diet.this, "Food not found", Toast.LENGTH_LONG).show();
        } else {

            editText=(findViewById(R.id.editTextDiet));
            editText.setText(calories);
        }


    }
}

我有以下问题：

W/System.err: org.json.JSONException: No value for common
    at org.json.JSONObject.get(JSONObject.java:392)

所以问题似乎是“ common”数组没有值，因此找不到长度？我不确定为什么它看不到“普通”数组，因为我已经看过许多其他有关从数组中获取对象的问题，并且每次都相同地复制代码，但结果相同。如果我仅使用一个JSONObject并忽略了整个数组，那么我可以在stacktrace中看到它正在尝试将整个数组下载到该对象中，这意味着GET请求或API密钥绝对没有问题。谢谢。

Answer 1

您得到的错误是因为您没有将String响应传递给jsonObject，所以它无法在空对象中找到任何东西

解决方法是

String inputString;
while ((inputString = bufferedReader.readLine()) != null) {
         builder.append(inputString);
       }

        JSONObject jsonRes = new JSONObject(inputString); \\this is the fix 
        JSONArray common = jsonRes.getJSONArray("common");

无法从JSON数组获取JSON对象

1 个答案: