Splash Screen＆Premise Screen - Android

Question

所以我开始为一个简单的Android游戏制作一个简单的“启动画面”（它更像是一个介绍画面），我正在为一个移动编程课程做一个简单的安卓游戏。我遇到了问题。我知道我应该使用一个线程，但我的实现似乎不工作。你应该能够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(); 

} 

问题是，第一个textview与第一个wait（）调用一起被调用，但随后我得到一个强制关闭。

这里是我的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

这是我用我的闪屏代码：

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(); 

与闪屏的问题是，你使用的wait（）不正确。你在找什么是睡眠（）。

要触摸工作线程的UI线程，您需要使用runOnUiThread（Runnable action）或处理程序。

Answer 2

如果你想使用一个线程在这里，您必须调用start启动线程。这又将调用run（）方法：

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

splashThread.start(); 

请记住，这将在后台运行，即不是UI线程。

Answer 3

您需要使用Handler触摸UI（即设置你的textviews'文本。）

只是你的线程声明的上面，把这个：

Handler handler = new Handler(); 

然后而不是做first.setText("bla bla bla");

做

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

你需要这样做，因为单独的线程不能触摸UI元素。使用处理器基本上告诉UI线程运行任何你在传递

Answer 4

通过类似更换内部splashThread线程代码：

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

这样UI线程可以完美地触摸自己的看法。