【发布时间】:2014-04-15 11:14:18
【问题描述】:
我正在使用进程对话框功能来显示对话框。问题是对话框首先显示进程,然后进入运行方法,但我希望它在显示进程时移动到运行方法中。我怎样才能做到这一点...
AlertDialog.Builder ob=new AlertDialog.Builder(this);
ob.setTitle("Confrimation").setMessage("Are you sure you want to logout?");
ob.setNegativeButton("Cancel", new DialogInterface.OnClickListener(){
public void onClick(DialogInterface di1,int id)
{
dialog.cancel();
}
});
ob.setPositiveButton("OK", new DialogInterface.OnClickListener(){
public void onClick(DialogInterface di2,int id)
{
dialog = ProgressDialog.show(Questionnaire.this, "Processing","Downloading survey...");
Thread t=new Thread()
{
public void run()
{
// delete database values
deleteDatabaseValues();
Log.e("inside","dialogbox");
downloadDatabase();
dialog.dismiss();
}
};
Handler handler=new Handler();
handler.postDelayed(t,60000);
}
});
ob.show();
}
【问题讨论】:
-
使用了 AsynsTask 而不是 Thread 和
标签: android android-alertdialog dialog