Is it possible start a new thread within an Async task? Something like this:
是否可以在异步任务中启动新线程?像这样的东西:
public class FirstActivity extends Activity {
protected ProgressBar progBar;
protected Intent intent;
/** Called when the activity is first created. */
@Override
public void onCreate(Bundle savedInstanceState) {
super.onCreate(savedInstanceState);
setContentView(R.layout.main);
progBar = (ProgressBar)findViewById(R.id.start_progressBar);
progBar.setProgress(0);
new StartingApp().execute();
}
protected class StartingApp extends AsyncTask<Void, Integer, Void> {
int myProgress;
@Override
protected void onPreExecute() {
myProgress = 0;
}
@Override
protected Void doInBackground(Void... params) {
while(myProgress<50){
myProgress++;
publishProgress(myProgress);
SystemClock.sleep(10);
}
MyRunnableClass mrc = new MyRunnableClass();
mrc.run();
return null;
}
@Override
protected void onPostExecute(Void result){
intent = new Intent(FirstActivity.this, SecondActivity.class);
startActivity(intent);
}
@Override
protected void onProgressUpdate(Integer... values) {
progBar.setProgress(values[0]);
}
}
}
MyRunnableClass is a class which implements Runnable. I want something like this because in the first activity I want to show a progress bar while the application is initializing (fill data structures, starting threads).
MyRunnableClass是一个实现Runnable的类。我想要这样的东西,因为在第一个活动中我想在应用程序初始化时显示进度条(填充数据结构,启动线程)。
Another question I have is: should I use the run() or start() method?
我的另一个问题是:我应该使用run()或start()方法吗?
Thanks in advance!
提前致谢!
1 个解决方案
#1
0
Why do you want to do that? As pointed out in the code you need to call new Thread(mrc).start() to make it work. Otherwise i dont see any problem in that code spawning a new thread.
你为什么要那样做?正如在代码中指出的那样,您需要调用新的Thread(mrc).start()来使其工作。否则我没有看到产生新线程的代码有任何问题。
#1
0
Why do you want to do that? As pointed out in the code you need to call new Thread(mrc).start() to make it work. Otherwise i dont see any problem in that code spawning a new thread.
你为什么要那样做?正如在代码中指出的那样,您需要调用新的Thread(mrc).start()来使其工作。否则我没有看到产生新线程的代码有任何问题。