Java NullPointerException，简单的错误？

我不太清楚为什么我会得到一个NullPointerException，这是Java的新功能，因此非常感谢您的帮助

public class MainActivity extends AppCompatActivity {


    private TextView httpstuff;

    @Override
    protected void onCreate(Bundle savedInstanceState) {
        super.onCreate(savedInstanceState);
        setContentView(R.layout.activity_main);
        TextView httpstuff = (TextView)findViewById(R.id.tvHttp);
        Button button = (Button) findViewById(R.id.button);
        button.setOnClickListener(
            new Button.OnClickListener(){
                public void onClick(View v){
                    new JSONTask().onPostExecute("http://samples.openweathermap.org/data/2.5/weather?q=Chicago&appid=93eea9a9277a0520d2f444c3ff8c62da");
                }
            }
    );

}

   class JSONTask extends AsyncTask<String,String,String> {
        @Override
        protected String doInBackground(String... urls) {
        HttpsURLConnection connection = null;
        BufferedReader reader = null;
        try {
            URL url = new URL("https://samples.openweathermap.org/data/2.5/weather?q=Chicago&appid=93eea9a9277a0520d2f444c3ff8c62da");
            connection = (HttpsURLConnection) 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);
       httpstuff.setText(result);
        }
    }
}

在

new JSONTask().onPostExecute("http://samples.openweathermap.org/data/2.5/weather?q=Chicago&appid=93eea9a9277a0520d2f444c3ff8c62da");


private TextView httpstuff is a designated as a field but says that it is not initialized, but I use it in class JSONTask? Any ideas on why the exception is occurring? Thanks alot in advance!  

删除第一个文本视图

在学习时始终尝试使用

this.httpstuff

（或

main activity.this.httpstuff

），直到您了解何时不使用

this

httpstuff被指定为一个字段，但表示它未初始化

您正在Asynctask中使用该字段，是的，但您从未初始化它，正如警告所说

---

OnCreate有一个同名的局部变量

您正在跟踪

httpstuff

谢谢您的帮助！我删除了onCreate中的第一个textview，但仍然在同一行中得到异常，如果我遗漏了一些明显的内容，很抱歉。findViewById返回null。我认为不能将文本设置为空字符串。您可以在代码中插入print语句，以查看使用View.onClickListener（）会发生什么情况来解决此问题？您已经在使用它了。Button.onClickListener实际上并不存在，我现在正在看一个教程，我将代码一行一行地复制，但仍然会出错..hmm

new JSONTask().onPostExecute("http://samples.openweathermap.org/data/2.5/weather?q=Chicago&appid=93eea9a9277a0520d2f444c3ff8c62da");


private TextView httpstuff is a designated as a field but says that it is not initialized, but I use it in class JSONTask? Any ideas on why the exception is occurring? Thanks alot in advance!  

 TextView httpstuff = (TextView)