Splash Screen＆Premise Screen - Android

Question

所以我開始爲一個簡單的Android遊戲製作一個簡單的「啓動畫面」（它更像是一個介紹畫面），我正在爲一個移動編程課程做一個簡單的安卓遊戲。我遇到了問題。我知道我應該使用一個線程，但我的實現似乎不工作。你應該能夠感受到我要去的效果。

public class SplashScreen extends Activity { 

//how long until we go to the next activity 
protected int splashTime = 2000; 
private Thread splashThread; 

public void onCreate(Bundle savedInstanceState) { 
    super.onCreate(savedInstanceState); 
    setContentView(R.layout.splash_screen); 

    final TextView first = (TextView) findViewById(R.id.textView1); 
    final TextView second = (TextView) findViewById(R.id.textView2); 
    final TextView third = (TextView) findViewById(R.id.textView3); 
    final TextView fourth = (TextView) findViewById(R.id.textView4); 
    final TextView fifth = (TextView) findViewById(R.id.textView5); 
    final TextView sixth = (TextView) findViewById(R.id.textView6); 
    final ImageView main_character = (ImageView) findViewById(R.id.imageView1); 

    // thread for displaying the SplashScreen 
    splashThread = new Thread() { 
     @Override 
     public void run() { 
      try { 
       synchronized(this){ 

        first.setText("The year is 2048..."); 

        wait(splashTime); 

        first.setText(""); 
        second.setText("...the Earth's resources have long been depleted"); 

        wait(splashTime); 

        second.setText(""); 
        third.setText("Attempts have been made to save our home..."); 

        wait(splashTime); 

        third.setText(""); 
        fourth.setText("...but all has gone awry, trash now rains from the skies"); 

        wait(splashTime); 

        fourth.setText(""); 
        fifth.setText("Now, only one man can save us, and his name is..."); 

        wait(splashTime); 

        fifth.setText(""); 
        sixth.setText("The Garbage Man!"); 
        main_character.setImageResource(drawable.bobrgb888); 

       } 
      } catch (InterruptedException e) { 
       e.printStackTrace(); 
      } 
      finally { 
       finish(); 

       //start a new activity 
       Intent i = new Intent(getApplicationContext(), MainMenu.class); 
       startActivity(i); 
      } 
     } 
    }; 
    splashThread.start(); 

} 

問題是，第一個textview與第一個wait（）調用一起被調用，但隨後我得到一個強制關閉。

這裏是我的logcat：

07-02 19:35:16.241: W/dalvikvm(709): threadid=9: thread exiting with uncaught exception (group=0x40015560) 
07-02 19:35:16.250: E/AndroidRuntime(709): FATAL EXCEPTION: Thread-10 
07-02 19:35:16.250: E/AndroidRuntime(709):  android.view.ViewRoot$CalledFromWrongThreadException: Only the original thread that created a view hierarchy can touch its views. 
07-02 19:35:16.250: E/AndroidRuntime(709): at android.view.ViewRoot.checkThread(ViewRoot.java:2932) 
07-02 19:35:16.250: E/AndroidRuntime(709): at android.view.ViewRoot.requestLayout(ViewRoot.java:629) 
07-02 19:35:16.250: E/AndroidRuntime(709): at android.view.View.requestLayout(View.java:8267) 
07-02 19:35:16.250: E/AndroidRuntime(709): at android.view.View.requestLayout(View.java:8267) 
07-02 19:35:16.250: E/AndroidRuntime(709): at android.view.View.requestLayout(View.java:8267) 
07-02 19:35:16.250: E/AndroidRuntime(709): at android.view.View.requestLayout(View.java:8267) 
07-02 19:35:16.250: E/AndroidRuntime(709): at android.widget.RelativeLayout.requestLayout(RelativeLayout.java:257) 
07-02 19:35:16.250: E/AndroidRuntime(709): at android.view.View.requestLayout(View.java:8267) 
07-02 19:35:16.250: E/AndroidRuntime(709): at android.widget.TextView.checkForRelayout(TextView.java:5521) 
07-02 19:35:16.250: E/AndroidRuntime(709): at android.widget.TextView.setText(TextView.java:2724) 
07-02 19:35:16.250: E/AndroidRuntime(709): at android.widget.TextView.setText(TextView.java:2592) 
07-02 19:35:16.250: E/AndroidRuntime(709): at android.widget.TextView.setText(TextView.java:2567) 
07-02 19:35:16.250: E/AndroidRuntime(709): at com.connor.black.SplashScreen$1.run(SplashScreen.java:41) 
07-02 19:35:16.420: D/szipinf(709): Initializing inflate state 

Answer 1

這是我用我的閃屏代碼：

final SplashScreen splash = this; 

Thread splashThread = new Thread() { 
     @Override 
     public void run() { 
      go = true; 
      while (go) { 
      try { 
       int waited = 0; 
       while(_active && (waited < _splashTime)) { 
        sleep(100); 
        if(_active) { 
         waited += 100; 
        } 
       } 
      } catch(InterruptedException e) { 
       // do nothing 
      } finally { 
       finish(); 
       Intent i = new Intent(); 
       i.setClass(splash, MyActivity.class); 
       startActivity(i); 
       //stop(); 
       go = false; 
      } 
      } 
     } 
    }; 
    splashThread.start(); 

與閃屏的問題是，你使用的wait（）不正確。你在找什麼是睡眠（）。

要觸摸工作線程的UI線程，您需要使用runOnUiThread（Runnable action）或處理程序。

Answer 2

如果你想使用一個線程在這裏，您必須調用start啓動線程。這又將調用run（）方法：

splashTread = new Thread() { 
    @Override 
    public void run() { 
     ... 
    } 
}; 

splashThread.start(); 

請記住，這將在後臺運行，即不是UI線程。

Answer 3

您需要使用Handler觸摸UI（即設置你的textviews'文本。）

只是你的線程聲明的上面，把這個：

Handler handler = new Handler(); 

然後而不是做first.setText("bla bla bla");

做

handler.post(new Runnable(){ 
    public void run(){ 
    first.setText("bla bla bla"); 
    } 
}); 

你需要這樣做，因爲單獨的線程不能觸摸UI元素。使用處理器基本上告訴UI線程運行任何你在傳遞

Answer 4

通過類似更換內部splashThread線程代碼：

runOnUiThread(new Runnable() { 
    public void run() { 
     first.setText("The year is 2048..."); 
     //... The rest of code goes here too. 
    } 
}); 

這樣UI線程可以完美地觸摸自己的看法。