Question

I had seen this error in this line: new JSONTask.execute("https://jsonparsingdemo-cec5b.firebaseapp.com/jsonData/moviesDemoItem.txt");

我正在开发一个应用程序，该应用程序创建URL连接，JSON解析，Asintask。但是我得到一个错误，给了我一张图像。

鉴于此错误无法解析符号执行。

公共类MainActivity扩展了AppCompatActivity { 私人TextView tvData;

@Override
protected void onCreate(Bundle savedInstanceState) {
    super.onCreate(savedInstanceState);
    setContentView(R.layout.activity_main);

    Button btnHit = findViewById(R.id.btnHit);
     tvData = findViewById(R.id.tvJsonItem);

    btnHit.setOnClickListener(new View.OnClickListener() {
        @Override
        public void onClick(View v) {
            new JSONTask.execute("https://jsonparsingdemo-cec5b.firebaseapp.com/jsonData/moviesDemoItem.txt");
        }
    });


}

public class JSONTask extends AsyncTask<String , String , String>{

    @Override
    protected String doInBackground(String... urls) {
        HttpURLConnection connection = null;
        BufferedReader reader = null;

        try {
            URL url = new URL(urls[0]);
            connection =(HttpURLConnection) url.openConnection();
            connection.connect();

            InputStream stream = connection.getInputStream();
            reader = new BufferedReader(new InputStreamReader(stream));
            StringBuffer  buffer = new StringBuffer();
            String line ="";
            while ((line = reader.readLine()) != null){
                buffer.append(line);
            }
            return buffer.toString();


        }catch (MalformedURLException e){
            e.printStackTrace();
        } catch (IOException e) {
            e.printStackTrace();
        }finally {
            if (connection != null) {
                connection.disconnect();
            }
            try {
                if (reader != null) {
                    reader.close();
                }
            } catch (IOException e) {
                e.printStackTrace();
            }
        }
        return null;
    }

    @Override
    protected void onPostExecute(String result) {
        super.onPostExecute(result);
        tvData.setText(result);
    }
  }
}

我想得到答案

Answer 1

使用此

new JSONTask().execute("https://jsonparsingdemo-cec5b.firebaseapp.com/jsonData/moviesDemoItem.txt");

代替这个

new JSONTask.execute("https://jsonparsingdemo-cec5b.firebaseapp.com/jsonData/moviesDemoItem.txt");

我该如何解决-无法解决符号执行

1 个答案: