它可以吗? 我有一个活动和一个警报对话。 但我需要先运行活动,然后2秒后出现alertdialog。 我不知道怎么做。问候
pd:我不是英语人士
public class Pantalladeinicio extends Activity {
private final int SPLASH_DISPLAY_LENGTH = 2000;
@Override
public void onCreate(Bundle savedInstanceState) {
super.onCreate(savedInstanceState);
requestWindowFeature(Window.FEATURE_NO_TITLE);
getWindow().setFlags(WindowManager.LayoutParams.FLAG_FULLSCREEN,
WindowManager.LayoutParams.FLAG_FULLSCREEN);
setContentView(R.layout.index);
if(checkInternetConnection()==true) {
new Handler().postDelayed(new Runnable() {
public void run() {
Intent mainIntent = new Intent(Pantalladeinicio.this,
NetworkingActivity.class);
mainIntent.putExtra("MAIN", true);
startActivity(mainIntent);
finish();
}
}, SPLASH_DISPLAY_LENGTH);
}
else
{
AlertDialog.Builder dialogo1 = new AlertDialog.Builder(this);
dialogo1.setTitle("Importante");
dialogo1.setMessage("Debe activar la Conexion a Internet");
dialogo1.setCancelable(false);
dialogo1.setPositiveButton("Aceptar", new DialogInterface.OnClickListener() {
public void onClick(DialogInterface dialogo1, int id) {
aceptar();
}
});
dialogo1.show();
Log.v("TAG", "Internet Connection Not Present");
}
}
public void aceptar() {
// Toast t=Toast.makeText(this,"Bienvenido a probar el programa.", Toast.LENGTH_SHORT);
super.onDestroy();
finish();
}
private boolean checkInternetConnection() {
ConnectivityManager cm = (ConnectivityManager) getSystemService(Context.CONNECTIVITY_SERVICE);
// test for connection
if (cm.getActiveNetworkInfo() != null
&& cm.getActiveNetworkInfo().isAvailable()
&& cm.getActiveNetworkInfo().isConnected()) {
return true;
} else {
//Log.v("TAG", "Internet Connection Not Present");
return false;
}
}
}
答案 0 :(得分:1)
我不明白你的问题,但看看接受的答案,我建议你改变现有代码的顺序:
new Handler().postDelayed(new Runnable() {
public void run() {
if(checkInternetConnection()) {
Intent mainIntent = new Intent(Pantalladeinicio.this, NetworkingActivity.class);
mainIntent.putExtra("MAIN", true);
startActivity(mainIntent);
finish();
}
else
{
AlertDialog.Builder dialogo1 = new AlertDialog.Builder(this);
dialogo1.setTitle("Importante");
dialogo1.setMessage("Debe activar la Conexion a Internet");
dialogo1.setCancelable(false);
dialogo1.setPositiveButton("Aceptar", new DialogInterface.OnClickListener() {
public void onClick(DialogInterface dialogo1, int id) {
aceptar();
}
});
dialogo1.show();
Log.v("TAG", "Internet Connection Not Present");
}
}
}, SPLASH_DISPLAY_LENGTH);
答案 1 :(得分:0)
将您的代码更改为使用Thread
和runOnUiThread
//Your Code here...
else
{
myThread();
}
}
public void myThread(){
Thread th=new Thread(){
@Override
public void run(){
try
{
Thread.sleep(2000);
Pantalladeinicio.this.runOnUiThread(new Runnable() {
@Override
public void run() {
// TODO Auto-generated method stub
AlertDialog.Builder dialogo1 = new AlertDialog.Builder(Pantalladeinicio.this);
dialogo1.setTitle("Importante");
dialogo1.setMessage("Debe activar la Conexion a Internet");
dialogo1.setCancelable(false);
dialogo1.setPositiveButton("Aceptar", new DialogInterface.OnClickListener() {
public void onClick(DialogInterface dialogo1, int id) {
aceptar();
}
});
dialogo1.show();
Log.v("TAG", "Internet Connection Not Present");
}
});
}catch (InterruptedException e) {
// TODO: handle exception
}
}
};
th.start();
}
答案 2 :(得分:0)
我假设你想进入else块后等待2秒钟。
您可以在致电Thread.sleep(2000);
之前致电dialogo1.show();
来执行此操作。但是,这会使您的程序显示为“冻结”。为了避免这种情况,你可以创建一个单独的线程来休眠,然后显示对话框。