How to use the random function to randomly select a string from an array?

[granite horizon]

Right now I'm trying to make a program to take user input and have the program take in five names and select three names. At the moment I have it taking in the five names and giving the input each a value to one of the subvalues of the array but I don't know how to randomly select three names because the random function only works with integers. Below is what I have so far (after I get this solved I will simplify the input portion with a nested loop lmao:

import java.util.Scanner;
import javax.swing.JOptionPane;
import java.util.Random;

public class Main {

public static void main(String[] args) {
    
    Scanner scanner = new Scanner(System.in);
    
    String[] names = new String[5];
    
    Random selecter = new Random();
    
    names[0] = JOptionPane.showInputDialog("Please enter a name (1/5): ");
    names[1] = JOptionPane.showInputDialog("Please enter a name (2/5): ");
    names[2] = JOptionPane.showInputDialog("Please enter a name (3/5): ");
    names[3] = JOptionPane.showInputDialog("Please enter a name (4/5): ");
    names[4] = JOptionPane.showInputDialog("Please enter a name (5/5): ");
    
    }

}

[calm glacier]

Hey, @granite horizon!
Please remember to /close this post once your question has been answered!

[finite stump]

As you've noticed, these strings are in an array, each in a different index, 0 to 4

So pick a random number from 0 to 4 and take the string that's at that index

[granite horizon]

What would be the code to specifically call those 5 numbers?

[finite stump]

Those 5 numbers are 0 to 4

It's the same as for any range of numbers from 0 to something

[granite horizon]

Oh, so I don't have to specify that they're integers in an array?

[finite stump]

Not when you pick them, no

You'll notice that computers tend to royally not care why you make them do stuff

[granite horizon]

I'm sorry, I'm a little new - is this what you're wanting me to create?
int randomizer = selecter(5);

@finite stump ^

[finite stump]

I want you to use Random in the normal way

[granite horizon]

What is the normal way?

[finite stump]

Do you just refuse to look it up?

[granite horizon]

.

When I look up "creating a random function in the normal way for java" it tells me how to create the random class and then how to have it randomly select numbers in a specific order

[granite horizon]

What I'm asking is how do I take the output of the randomly selected number in the random class and assign it to the 2d arrays

Because the random class will give me, let's say a 4, but it will just be an integer, it won't output the 4th subvalue on my array

[granite horizon]

@radiant knoll

[radiant knoll]

is it ok to select the same name multiple times

@granite horizon

[granite horizon]

Sure

[radiant knoll]

did you see this https://www.baeldung.com/java-random-list-element#1-single-random-item

this one specifically uses an array so you don't have to change list semantics into array semantics https://stackoverflow.com/a/8065554/3243226

[granite horizon]

Oh awesome, I'll check it out and let you know if it works, thanks!!

[radiant knoll]

no worries

[granite horizon]

So I have a couple of questions

When I use this piece of code:

public static int getRandom(int[] array) {
int rnd = new Random().nextInt(array.length);
return array[rnd];
}
Am I using the name of my array in the "array" section? Because when I put the name of my array it's giving me the error of "names can't be resolved to a variable"

Because the whole point was I'm trying to randomly select a string variable and the link attached is just with integers

[radiant knoll]

Show code

[radiant knoll]

I assume you've just used the method but it can't access your array

So you can just move the code into your main class or give the array to the method as a parameter

[granite horizon]

Random generator = new Random(); int randomIndex = generator.nextInt(names.length); return names[randomIndex];

Here is the full code:
`import java.util.Scanner;
import javax.swing.JOptionPane;
import java.util.Random;

public class Main {

public static void main(String[] args) {
    
    Scanner scanner = new Scanner(System.in);
    
    String[] names = new String[5];
    
    names[0] = JOptionPane.showInputDialog("Please enter a name (1/5): ");
    names[1] = JOptionPane.showInputDialog("Please enter a name (2/5): ");
    names[2] = JOptionPane.showInputDialog("Please enter a name (3/5): ");
    names[3] = JOptionPane.showInputDialog("Please enter a name (4/5): ");
    names[4] = JOptionPane.showInputDialog("Please enter a name (5/5): ");
    
    scanner.close();
    
    System.out.println(rnd);
    
    Random generator = new Random();
    int randomIndex = generator.nextInt(names.length);
    return names[randomIndex];
    
    }

}`

[radiant knoll]

yeah you just need String myRandomName = names[randomIndex];

you're not returning anything here

@granite horizon

[granite horizon]

Oh okay

So don't make it an int?

Oh wait hold on

[granite horizon]

Okay here's what I changed it to:
` Random generator = new Random();
String randomIndex = names[randomIndex];

    System.out.println(generator);`

[radiant knoll]

you want to print randomIndex not generator

[granite horizon]

But I'm getting an error under the randomIndex portion saying that it's going from int to string

[radiant knoll]

wait

you've compressed 3 lines into 2 for some reason

[granite horizon]

I removed the return line

[radiant knoll]

don't remove it, changei t

[granite horizon]

Ohhh

[radiant knoll]

        Random generator = new Random();
        int randomIndex = generator.nextInt(names.length);
        String randomlySelectedName = names[randomIndex];

[granite horizon]

So randomIndex is an int?

[radiant knoll]

yes, hence int randomIndex

[granite horizon]

Interesting, okay one sec

[radiant knoll]

but instead of returning it (to nowhere), save it to a variable called - in this case - randomlySelectedName

[granite horizon]

And am I printing randomlySelectedName or randomIndex

[radiant knoll]

which one do you think

[granite horizon]

randomlySelectedName

Because it's the final string that results all of names and randomIndex?

[radiant knoll]

it's the randomly selected name

so if you want to print that ou then yeah it's in that variable

[granite horizon]

Oh wait

😂

[radiant knoll]

why not just try it

give it a go and see what happens, print both and see what each does

[granite horizon]

It works!!

[radiant knoll]

the computer won't be upset you ran the program multiple times

[granite horizon]

Okay so I understand most of this but I want to be able 100% of it so I don't ask any stupid questions for next time, in the code below:
`import java.util.Scanner;
import javax.swing.JOptionPane;
import java.util.Random;

public class Main {

public static void main(String[] args) {
    
    Scanner scanner = new Scanner(System.in);
    
    String[] names = new String[5];
    
    names[0] = JOptionPane.showInputDialog("Please enter a name (1/5): ");
    names[1] = JOptionPane.showInputDialog("Please enter a name (2/5): ");
    names[2] = JOptionPane.showInputDialog("Please enter a name (3/5): ");
    names[3] = JOptionPane.showInputDialog("Please enter a name (4/5): ");
    names[4] = JOptionPane.showInputDialog("Please enter a name (5/5): ");
    
    scanner.close();
    
    Random generator = new Random();
    int randomIndex = generator.nextInt(names.length);
    String randomlySelectedName = names[randomIndex];
    
    System.out.println(randomlySelectedName);
    
    }

}`

What is specifically happening that's bypassing the random function of needing an integer input and allowing it to use a string input

Line 21 I'm creating a random function, but line 22 is where I slightly get lost

(Line 22 is int randomIndex = generator.nextInt(names.length);)

I'm sorry for all the questions haha 😅

[radiant knoll]

What is specifically happening that's bypassing the random function of needing an integer input

nothing, we're using the array's length as the (integer) input

we're passing names.length as the parameter to nextInt; names.length is an int

[granite horizon]

Ohhhh

So arrays already have an integer input and instead of reading the string function we're just reading the position of the string function

[radiant knoll]

arrays report their length using .length, this value is an integer; is that clear so far

[granite horizon]

Yup

[radiant knoll]

generator.nextInt(...) takes an int argument, do you agree?

[granite horizon]

Yes

[radiant knoll]

so if the array produces its length as an int, and nextInt takes an int, i can pass the array's length into nextInt

[granite horizon]

Because arrays have positions

[radiant knoll]

that's the next step

[granite horizon]

Like in a 1D array, the position of the 3rd input would be 2

[radiant knoll]

yes but that's the next step, i need to ask for a randomly generated position

[granite horizon]

Oh okay

[radiant knoll]

but in order to get Random to not generate any old random shite, i can give it an upper bound (which is the int argument)

so what i'm saying to random is "please generate me a random number between 0 and this int input (exclusive)"

[granite horizon]

Okay

[radiant knoll]

the int input is the length of this array

random says ok here you go....2

i now have the random position generated for me by Random

i can now go back to my array and say hey can you give me the 2th position please

it says here you go, jeff

so step 1 is ask random for a random number up to (but not including) the length of my array
step 2 is ask the array for the array position corresponding to taht randomly generated number

[granite horizon]

Okay, I understand that

[radiant knoll]

so that's how we randomly select a value from an array

it might be useful to look at the docs of nextInt(int):

public int nextInt(int bound)

Returns a pseudorandom, uniformly distributed int value between 0 (inclusive) and the specified value (exclusive),

source: https://docs.oracle.com/en/java/javase/17/docs/api/java.base/java/util/Random.html#nextInt(int)

[granite horizon]

Okay cool, thank you so much!!!

[radiant knoll]

no worries