具有 json 的应用程序-在非 json 数据后简单崩溃

Question

我有

 HttpURLConnection urlConnection = null;
        String result = "";
        try {

            String host = "http://www.example.com/json.json";

            URL url = new URL(host);
            urlConnection = (HttpURLConnection) url.openConnection();

            int code = urlConnection.getResponseCode();

            if(code==200){
                InputStream in = new BufferedInputStream(urlConnection.getInputStream());
                if (in != null) {
                    JSONParser jsonParser = new JSONParser();
                    JSONObject jsonObject = (JSONObject)jsonParser.parse(new InputStreamReader(in, "UTF-8"));
                    result=(String) jsonObject.get("name");
                    System.out.print(jsonObject);
                }
                in.close();
            } else { result="9";}

            return result;
        } catch (MalformedURLException e) {
            result="9";
        } catch (IOException e) {
            result="9";
        }
        catch (ParseException e) {
            e.printStackTrace();
            result="9";
        }

        finally {
            urlConnection.disconnect();
        }
        return result;

当我输入有效的 json 数据时，一切正常，但如果我得到非 json 数据，我会遇到应用程序崩溃：

Caused by: java.lang.ClassCastException: java.lang.Long cannot be cast to org.json.simple.JSONObject

我认为

 catch (ParseException e) {
                e.printStackTrace();
                result="9";
            }

应该处理这个，但是没有。

所以我必须做些什么来避免应用程序在我没有获得有效时崩溃的情况json？

Answer 1

抛出的异常是ClassCastException。也许您也可以通过添加另一个 catch 来捕获该异常？

catch (ClassCastException e) {
    e.printStackTrace();
    result="9";
}

Answer 2

试试这个来发出 http 请求

class SendPostReqAsyncTask extends AsyncTask<String, Void, String> {

        @Override
        protected void onPreExecute() {
            super.onPreExecute();
        }

        @Override
        protected String doInBackground(String... params) {
            URL myUrl = null;
            HttpURLConnection conn = null;
            String response = "";
            //String data = params[0];

            try {
                myUrl = new URL("http://www.example.com/json.json");
                conn = (HttpURLConnection) myUrl.openConnection();
                conn.setReadTimeout(10000);
                conn.setConnectTimeout(15000);
                conn.setRequestMethod("POST");
                conn.setDoInput(true);
                conn.setDoOutput(true);

                //one long string, first encode is the key to get the  data on your web
                //page, second encode is the value, keep concatenating key and value.
                //theres another ways which easier then this long string in case you are
                //posting a lot of info, look it up.
                String postData = URLEncoder.encode("key", "UTF-8") + "=" +
                        URLEncoder.encode("value", "UTF-8");
                OutputStream os = conn.getOutputStream();

                BufferedWriter bufferedWriter = new BufferedWriter(new OutputStreamWriter(os, "UTF-8"));
                bufferedWriter.write(postData);
                bufferedWriter.flush();
                bufferedWriter.close();

                InputStream inputStream = conn.getInputStream();
                BufferedReader bufferedReader = new BufferedReader(new InputStreamReader(inputStream, "UTF-8"));
                String line = "";
                while ((line = bufferedReader.readLine()) != null) {
                    response += line;
                }
                bufferedReader.close();
                inputStream.close();
                conn.disconnect();
                os.close();
            } catch (MalformedURLException e) {
                e.printStackTrace();
            } catch (UnsupportedEncodingException e) {
                e.printStackTrace();
            } catch (ProtocolException e) {
                e.printStackTrace();
            } catch (IOException e) {
                e.printStackTrace();
            }

            return response;
        }

        @Override
        protected void onPostExecute(String s) {
            try {
                JSONObject jsonObject = new JSONObject(s);
            } catch (JSONException e) {
                //s may not be json
            }
        }
    }

Answer 3

在从 Json 对象获取字符串之前，检查 Json 对象是否不为 null 并且具有该字符串。然后尝试得到它。

if (jsonObject!=null && jsonObject.has("name"))
 {
     result = jsonObject.get("name");
     System.out.print(result);
  }

具有 json 的应用程序-在非 json 数据后简单崩溃

Aplication with json-simple crash after non json data

java

android

json

json-simple