Use the Random.nextInt(int) method:
final String[] proper_noun = {"Fred", "Jane", "Richard Nixon", "Miss America"};
Random random = new Random();
int index = random.nextInt(proper_noun.length);
System.out.println(proper_noun[index]);
This code is not completely safe: one time out of four it'll choose Richard Nixon.
To quote a documentation Random.nextInt(int):
Returns a pseudorandom, uniformly distributed int value between 0 (inclusive) and the specified value (exclusive)
In your case passing an array length to the nextInt will do the trick - you'll get the random array index in the range [0; your_array.length)
Use the Random.nextInt(int) method:
final String[] proper_noun = {"Fred", "Jane", "Richard Nixon", "Miss America"};
Random random = new Random();
int index = random.nextInt(proper_noun.length);
System.out.println(proper_noun[index]);
This code is not completely safe: one time out of four it'll choose Richard Nixon.
To quote a documentation Random.nextInt(int):
Returns a pseudorandom, uniformly distributed int value between 0 (inclusive) and the specified value (exclusive)
In your case passing an array length to the nextInt will do the trick - you'll get the random array index in the range [0; your_array.length)
if you use List instead of arrays you can create simple generic method which get you random element from any list:
public static <T> T getRandom(List<T> list)
{
Random random = new Random();
return list.get(random.nextInt(list.size()));
}
if you want to stay with arrays, you can still have your generic method, but it will looks bit different
public static <T> T getRandom(T[] list)
{
Random random = new Random();
return list[random.nextInt(list.length)];
}
Videos
First you'll have to move the S array to be an instance variable or a static variable, because currently it's local to your main method, and can't be accessed from your get method.
Then you can get a random String this way :
private Random rnd = new Random();
public String get ()
{
return S[rnd.nextInt(S.length)];
}
You can use java.util.Random to generate random things. However, keep in mind that, it is not secure, not really random.
You can get random chars from the array S:
String randomString = "";
Random rand = new Random();
for(int i=0;i<=4;i++)
{
randomString += S[rand.nextInt(S.length())].charAt(0);
}
System.out.println(randomString);
Two things you have to change:
- currentRoom should be String Array
Printing string array should be done using Arrays.toString
String[] currentRoom; String [][] rooms = {{"Start", "Treasure Room1"}, {"Goblin Home1", "Spider Nest1"}}; Random rand = new Random(); currentRoom = rooms [rand.nextInt(rooms.length)]; System.out.println(Arrays.toString(currentRoom));
Sample Output
[Goblin Home1, Spider Nest1]
Hope this helps!
Why do you need a 2D array of strings? I think you can manage with a 1D array instead.
String currentRoom;
String[] rooms = {"Start", "Treasure Room1", "Goblin Home1", "Spider Nest1"};
Random rand = new Random();
currentRoom = rooms [rand.nextInt( rooms.length)];
System.out.println(currentRoom);
Try:
list[r.nextInt(list.length)];
The accepted answers is not working for me the solution worked for me is
List<String> myList = Arrays.asList("A", "B", "C", "D");
Suppose you have this above ArrayList and you want to randomize it
Random r = new Random();
int randomitem = r.nextInt(myList.size());
String randomElement = myList.get(randomitem);
If you print this randomElement variable you will get random string from your ArrayList
Do you mean generating random string from this set of characters? You can perform it in this way:
int len = ...; // length of resulting string
StringBuilder builder = new StringBuilder();
Random rand = new Random();
for (int i = 0; i < len; ++i) {
String c = arry[rand.nextInt() % arry.length];
builder.append(c);
}
String str = builder.toString();
To get a String of characters you can do
static final String chars =
"ABCDEFGHIJKLMNOPQRSTUVWXYZ" +
"abcdefghijklmnopqrstuvwxyz" +
"0123456789`~!@#$%^&*()-_" +
"=+;:'|\"<,>./?";
static final Random rand = new Random();
public static String randString(int length) {
char[] gen = new char[length];
for (int i = 0; i < length; i++)
gen[i] = chars.charAt(rand.nextInt(chars.length));
return new String(gen);
}