Új hozzászólás Aktív témák
-
lanszelot
addikt
válasz
axioma
#12082
üzenetére
Nagyon szépen köszönöm a segítseget.
Sajnos nincs semmi ami ismétlődne.
public class MainActivity extends AppCompatActivity {
@Override
protected void onCreate(Bundle savedInstanceState) {
super.onCreate(savedInstanceState);
EdgeToEdge.enable(this);
setContentView(R.layout.activity_main);
ViewCompat.setOnApplyWindowInsetsListener(findViewById(R.id.main), (v, insets) -> {
Insets systemBars = insets.getInsets(WindowInsetsCompat.Type.systemBars());
v.setPadding(systemBars.left, systemBars.top, systemBars.right, systemBars.bottom);
return insets;
});
Random random = new Random();
ImageView imageView1 = findViewById(R.id.Egyes);
ImageView imageView2 = findViewById(R.id.Kettes);
ImageView imageView3 = findViewById(R.id.Harmas);
ImageView imageView4 = findViewById(R.id.Negyes);
ImageView imageView5 = findViewById(R.id.Otos);
Button button01 = findViewById(R.id.Dobas);
button01.setOnClickListener(new View.OnClickListener() {
@Override
public void onClick(View v) {
int[] kockak = new int[5];
//elso kocka
int randSzam = random.nextInt(5);
kockak[0] = randSzam;
switch(randSzam) {
case 0:
imageView1.setImageResource(R.drawable.one);
break;
case 1:
imageView1.setImageResource(R.drawable.two);
break;
case 2:
imageView1.setImageResource(R.drawable.three);
break;
case 3:
imageView1.setImageResource(R.drawable.four);
break;
case 4:
imageView1.setImageResource(R.drawable.five);
break;
case 5:
imageView1.setImageResource(R.drawable.six);
break;
}
//masodik kocka
randSzam = random.nextInt(5);
kockak[1] = randSzam;
switch(randSzam) {
case 0:
imageView2.setImageResource(R.drawable.one);
break;
case 1:
imageView2.setImageResource(R.drawable.two);
break;
case 2:
imageView2.setImageResource(R.drawable.three);
break;
case 3:
imageView2.setImageResource(R.drawable.four);
break;
case 4:
imageView2.setImageResource(R.drawable.five);
break;
case 5:
imageView2.setImageResource(R.drawable.six);
break;
}
//harmadik kocka
randSzam = random.nextInt(5);
kockak[2] = randSzam;
switch(randSzam) {
case 0:
imageView3.setImageResource(R.drawable.one);
break;
case 1:
imageView3.setImageResource(R.drawable.two);
break;
case 2:
imageView3.setImageResource(R.drawable.three);
break;
case 3:
imageView3.setImageResource(R.drawable.four);
break;
case 4:
imageView3.setImageResource(R.drawable.five);
break;
case 5:
imageView3.setImageResource(R.drawable.six);
break;
}
//negyedik kocka
randSzam = random.nextInt(5);
kockak[3] = randSzam;
switch(randSzam) {
case 0:
imageView4.setImageResource(R.drawable.one);
break;
case 1:
imageView4.setImageResource(R.drawable.two);
break;
case 2:
imageView4.setImageResource(R.drawable.three);
break;
case 3:
imageView4.setImageResource(R.drawable.four);
break;
case 4:
imageView4.setImageResource(R.drawable.five);
break;
case 5:
imageView4.setImageResource(R.drawable.six);
break;
}
//otodik kocka
randSzam = random.nextInt(5);
kockak[4] = randSzam;
switch(randSzam) {
case 0:
imageView5.setImageResource(R.drawable.one);
break;
case 1:
imageView5.setImageResource(R.drawable.two);
break;
case 2:
imageView5.setImageResource(R.drawable.three);
break;
case 3:
imageView5.setImageResource(R.drawable.four);
break;
case 4:
imageView5.setImageResource(R.drawable.five);
break;
case 5:
imageView5.setImageResource(R.drawable.six);
break;
}
}
});
}
}Másik fórumon ezt írták, de egy szót sem értek belőle /nem az angol miatt/
"Do you want to have a dynamic drawable resource ID? If yes, there is a way to get the ID by the resource name. However, I strongly suggest you think carefully before taking this path. Having a map where you associate static resource IDs with another value might be better."
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