將字符串轉換爲int在服務

Question

我試圖將String轉換爲integer。 Integer.parseInt()之前工作過，但現在它失敗了。這是在這裏失敗int INTERVAL= (60000 * Integer.parseInt(preferenceTime));我想動態地指定時間，當我安排一個計時器。
謝謝

public class Service extends Service { 

    public SharedPreferences settings; 


    private Handler HandleIt = new Handler(); 
    private Timer timer = new Timer(); 
    boolean timeout = false; 
    //private PowerManager pm = (PowerManager)getSystemService(POWER_SERVICE); 





    @Override 
    public IBinder onBind(Intent intent) { 
     // TODO Auto-generated method stub 
     return null; 
    } 

    ////////////////////////////////////////////// 

    class TimeDisplayTimerTask extends TimerTask { 



     @Override 
     public void run() { 
      HandleIt.post(new Runnable(){ 
       public void run(){ 

        //SharedPreferences 
        settings = getSharedPreferences("timer_preference", MODE_PRIVATE); 
        String preferenceTime = settings.getString("timer_preference", ""); 

        // int INTERVAL= (60000 * Integer.parseInt(preferenceTime)); 


        Toast.makeText(getApplicationContext(), TextonScreen(), Toast.LENGTH_SHORT).show(); 

        //get screen light up 
        PowerManager pm = (PowerManager)getSystemService(Context.POWER_SERVICE); 
        boolean isScreenOn = pm.isScreenOn(); 
        if(isScreenOn==false) { 
         pm.newWakeLock(PowerManager.SCREEN_DIM_WAKE_LOCK | PowerManager.ON_AFTER_RELEASE, "My Tag"); 
        } 
        // make a new intent and start it with flag and send an sms 
        Intent launch = new Intent(getBaseContext(), SMS.class); 
        launch.addFlags(Intent.FLAG_ACTIVITY_NEW_TASK); 
        startActivity(launch); 
       } 
      }); 
     } 


    } 

    private String TextonScreen() 
    { 
     timeout = true; 
     return "it is running"; 


    } 
    boolean isTimeout() 
    { 
     return timeout; 
    } 
    @Override 
    public void onCreate() { 

     // TODO Auto-generated method stub 
     super.onCreate(); 
     Toast.makeText(this, "Service is created", Toast.LENGTH_SHORT).show(); 



    } 

    @Override 
    public int onStartCommand(Intent intent, int flags, int startId) { 


     SharedPreferences settings = getSharedPreferences(getPackageName() + "timer_preference", MODE_PRIVATE); 
     String preferenceTime = settings.getString("timer_preference", ""); 
     int INTERVAL= (60000 * Integer.parseInt(preferenceTime)); 



     // TODO Auto-generated method stub 
     // Display the Toast Message 
     Toast.makeText(this, "Start Service", Toast.LENGTH_SHORT).show(); 
     // Execute an action after period time 
     //comes from the TimeDisplayTimerTask class 
     timer.scheduleAtFixedRate(new TimeDisplayTimerTask(), 0, INTERVAL); 


     return super.onStartCommand(intent, flags, startId); 
    } 


    @Override 
    public void onDestroy() { 
     // Display the Toast Message 
     Toast.makeText(this, "Stop Service", Toast.LENGTH_SHORT).show(); 
     if (timer != null) { 
      timer.cancel(); 
     } 
     super.onDestroy(); 
    } 

} 

Answer 1

Ofcourse它會失敗，因爲你想解析器「」作爲一個int！ 試試這個：

String preferenceTime = settings.getString("timer_preference", "0"); 
int INTERVAL= (60000 * Integer.parseInt(preferenceTime)); 

0意味着，如果我從來沒有在任何東西，返回0。它的默認值。

爲什麼不把它存儲爲int？！我的意思是這樣的：

int preferenceTime = settings.getInteger("timer_preference", 0); 
int INTERVAL= (60000 * preferenceTime); 

Answer 2

如果它不能解析一個數字，它會拋出一個NumberFormatException異常。我平時喜歡在Optional包Integer#parseInt電話，所以我可以提供一個默認值：

public static Optional<Integer> parseInt(String s) { 
    try { 
     return Optional.of(Integer.parseInt(s)); 
    } catch (NumberFormatException ex) { 
     return Optional.empty(); 
    } 
} 

然後，稱可以提供一個默認的（或者讓你知道它通過#isPresent返回false失敗）：

int value = parseInt(preferenceTime).orElse(/* some default integer */);