Random element from int to String?

Question

Okay, so I'll first show my code, then I'll explain what I need help with.

  Random isServerChance = new Random();
  Random isGood = new Random();
  Random isBad = new Random();

  // Some IDs which are considered "Good"
  int isGoodTable[] =
  { 1038, 1040, 1042, 1044 };
  // Some IDs which are considered "Bad"
  int isBadTable[] =
  { 6570, 6585, 952, 969 };
  // Based on this roll, will pick which cat is used. Good, or Bad.
  int isServerRoll = isServerChance.nextInt(6);
  // If it's > 3 it's going to be a "Good" item, if it's not, it's a bad
  // item.
  if (isServerRoll > 3)
  {

     int isGoodValue = isGood.nextInt(isGoodTable.length);
     System.out.println("You've received a rare item! You got a: " + isGoodTable[isGoodValue]);

  } else
  {
     int isBadValue = isBad.nextInt(isBadTable.length);
     System.out.println("You've received a normal item. You've got a: " + isBadTable[isBadValue]);

  }

}

Okay, so this will print: You've received a normal item. You've got a: 969 || You've received a rare item! You got a: 1042.

Let's say the int value of 969 is a Spade. Is it possible for me to make 969 equal to an array of Strings from one of the int tables above?

So, if I land on 969, it'll say: "You've received a normal item. You've got a: Spade!"

I'm sort of new, so I'm not sure if this is even possible, but I don't want to make loads of if statements for each ID.

Thanks

Answer 1

It seems that you simply want arrays of Strings rather than arrays of integers:

String[] isBadTable = { "Foo", "Bar", "Baz", "Spade" };

Read the tutorial on arrays.

Answer 2

Yes. Use a Map<Integer, String>. You could put your items in like this

Map<Integer, String> itemMap = new HashMap<Integer, String>();
itemMap.put(969, "Spade");
// etc.

Then you can retrieve your values like this

String itemName = itemMap.get(969);  // spade

If you do this, just be careful, because a HashMap doesn't have any default value, so if the key you try to use isn't in the Map you'll get null back.