android - 执行异步任务时应用程序崩溃

Question

我的应用程序在执行异步任务时崩溃。我尝试将数据从 android 发送到服务器，但我的应用程序在执行 AsyncTask 时崩溃 ` 输入 input = new input(); 输入.执行();

    mainActivity = this;
    latitude = "-6.711647";
    longitude ="108.5413";`


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

    HashMap<String, String> user = db.getUserDetails();
    String email = user.get("email");
    String success;

    @Override
    protected void onPreExecute() {
        super.onPreExecute();
        pDialog = new ProgressDialog(MainActivity.this);
        pDialog.setMessage("Sending Data to server...");
        pDialog.setIndeterminate(false);
        pDialog.show();
    }


    @Override
    protected String doInBackground(String... arg0) {
        String strEMAIL = email.toString();
        String strNama = latitude.toString();
        String strProdi = longitude.toString();


        List<NameValuePair> params = new ArrayList<NameValuePair>();
        params.add(new BasicNameValuePair("email", strEMAIL));
        params.add(new BasicNameValuePair("latitude", strNama));
        params.add(new BasicNameValuePair("longitude", strProdi));


        JSONObject json = jParser.makeHttpRequest(url, "POST", params);

        try {
            success = json.getString("success");

        } catch (Exception e) {
            Toast.makeText(getApplicationContext(), "Error",
                    Toast.LENGTH_LONG).show();
        }
        return null;
    }
    protected void onPostExecute(String file_url) {
        // dismiss the dialog once done
        pDialog.dismiss();

        if (success.equals("1")) 
        {
            Toast.makeText(getApplicationContext(), "kirim data Sukses!!!", Toast.LENGTH_LONG).show();
        } 
        else 
        {
            Toast.makeText(getApplicationContext(), "kirim data Gagal!!!", Toast.LENGTH_LONG).show();

        }
    }
}

这是我的完整代码 http://pastebin.com/jRdxeQKG 这是我的 logcat http://prntscr.com/7p3vbz

Answer 1

当您在onCreate() 中调用new input() 时，引用db 为null。

解决方案:在调用new input()之前初始化db即可:

 db = new SQLiteHandler(getApplicationContext());
 input input = new input();

顺便说一下，从设计的角度来看有很多问题，但这将解决(其中一个)崩溃。