无法在未调用 Looper.prepare() 的线程内创建处理程序

Question

我收到此错误“无法在未调用 Looper.prepare() 的线程内创建处理程序”

你能告诉我如何解决它吗？

public class PaymentActivity extends BaseActivity {
@Override
public void onCreate(Bundle savedInstanceState) {
    super.onCreate(savedInstanceState);
    setContentView(R.layout.payment);

    final Button buttonBank = (Button) findViewById(R.id.buttonBank);

    buttonBank.setOnClickListener(new View.OnClickListener() {
        public void onClick(View v) {
            progressDialog = ProgressDialog.show(PaymentActivity.this, "",
                    "Redirecting to payment gateway...", true, true);

            new Thread() {
                public void run() {
                    try {
                        startPayment("Bank");
                    } catch (Exception e) {
                        alertDialog.setMessage(e.getMessage());
                        handler.sendEmptyMessage(1);
                        progressDialog.cancel();
                    }
                }
            }.start();
        }

    });

开始付款方式：

    private void startPayment(String id) {
    Bundle b = getIntent().getExtras();
    final Sail sail = b.getParcelable(Constant.SAIL);

    final Intent bankIntent = new Intent(this, BankActivity.class);

    try {
        Reservation reservation = RestService.createReservation(
                sail.getId(),
                getSharedPreferences(Constant.PREF_NAME_CONTACT, 0));
        bankIntent.putExtra(Constant.RESERVATION, reservation);

        // <workingWithDB> Storing Reservation info in Database
        DBAdapter db = new DBAdapter(this);
        db.open();
        @SuppressWarnings("unused")
        long rowid;
        rowid = db.insertRow(sail.getId(), sail.getFrom(),
                sail.getTo(), sail.getShip(), sail.getDateFrom().getTime(),
                sail.getPrice().toString(), reservation.getId().floatValue());
        db.close();
        // </workingWithDB>

        String html = PaymentService.getRedirectHTML(id, reservation);

        bankIntent.putExtra(Constant.BANK, html);
    } catch (Exception e) {
        AlertDialog.Builder builder = new AlertDialog.Builder(this);
        AlertDialog alertDialog = builder.create();
        alertDialog.setMessage(e.getMessage());
        alertDialog.show();
    }

    startActivity(bankIntent);
}

Answer 1

你应该知道，当你尝试修改你的 UI 时，唯一线程可以做到这一点是UiThread。

所以如果你想在另一个线程中修改你的 UI，尝试使用方法：Activity.runOnUiThread(new Runnable);

你的代码应该是这样的：

 new Thread() {
    public void run() {  
        YourActivity.this.runOnUiThread(new Runnable(){

             @Override
             public void run(){
                 try {
                      startPayment("Bank");//Edit,integrate this on the runOnUiThread
                 } catch (Exception e) {
                     alertDialog.setMessage(e.getMessage());
                     handler.sendEmptyMessage(1);
                     progressDialog.cancel();
                 } 
            });                
           }
      }
  }.start();

Answer 2

我假设您在 startPayment() 方法中创建了一个处理程序。您不能这样做，因为只能在 UI 线程上创建处理程序。只需在您的活动类中创建它即可。

Answer 3

而不是new Thread() 行，尝试给予

this.runOnUiThread(new Runnable() {

Answer 4

您无法更改线程中的任何 UI，您可以使用 runOnUIThread 或 AsyncTask 了解有关此 click here 的更多详细信息

Answer 5

我发现大多数线程处理都可以用这样的 AsyncTasks 代替：

public class TestStuff extends Activity {
    @Override
    public void onCreate(Bundle savedInstanceState) {
        super.onCreate(savedInstanceState);
        setContentView(R.layout.main);

        Button buttonBank = (Button) findViewById(R.id.button);
        buttonBank.setOnClickListener(new View.OnClickListener() {
            public void onClick(View v) {
                new StartPaymentAsyncTask(TestStuff.this).execute((Void []) null);
            }
        });
    }

    private class StartPaymentAsyncTask extends AsyncTask<Void, Void, String> {
        private ProgressDialog dialog;
        private final Context context;

        public StartPaymentAsyncTask(Context context) {
            this.context = context;
        }

        @Override
        protected void onPreExecute() {
            dialog = new ProgressDialog(context);
            // setup your dialog here
            dialog.setProgressStyle(ProgressDialog.STYLE_SPINNER);
            dialog.setMessage(context.getString(R.string.doing_db_work));
            dialog.setCancelable(false);
            dialog.show();
        }

        @Override
        protected String doInBackground(Void... ignored) {
            String returnMessage = null;
            try {
                startPayment("Bank");
            } catch (Exception e) {
                returnMessage = e.getMessage();
            }
            return returnMessage;
        }

        @Override
        protected void onPostExecute(String message) {
            dialog.dismiss();
            if (message != null) {
                // process the error (show alert etc)
                Log.e("StartPaymentAsyncTask", String.format("I received an error: %s", message));
            } else {
                Log.i("StartPaymentAsyncTask", "No problems");
            }
        }
    }

    public void startPayment(String string) throws Exception {
        SystemClock.sleep(2000); // pause for 2 seconds for dialog
        Log.i("PaymentStuff", "I am pretending to do some work");
        throw new Exception("Oh dear, database error");
    }
}

我将应用程序上下文传递给异步，以便它可以从中创建对话框。

这样做的好处是您可以准确地知道哪些方法在您的 UI 中运行，哪些在单独的后台线程中。您的主 UI 线程没有延迟，并且分离成小的异步任务非常好。

代码假定您的 startPayment() 方法对 UI 没有任何作用，如果有，请将其移至 AsyncTask 的 onPostExecute 中，以便在 UI 线程中完成。

Answer 6

试试

final Handler handlerTimer = new Handler(Looper.getMainLooper());
        handlerTimer.postDelayed(new Runnable() {
            public void run() {
                                ...... 

                              }
                                                 }, time_interval});