Output Random Value From Array

Question

I have the following javascript code:

  let gemStones = [
  "Amethyst",
  "Diamond",
  "Emerald",
  "Ruby",
  "Sapphire",
  "Topaz",
  "Onyx",
];

let randomGemStone = gemStones[Math.floor(Math.random()*gemStones.length)];

function findGems()
{
  console.log("You found a " + randomGemStone + "!");
}

Here's the html:

<!DOCTYPE html>
<html lang="en">
<head>
  <meta charset="UTF-8">
  <meta http-equiv="X-UA-Compatible" content="IE=edge">
  <meta name="viewport" content="width=device-width, initial-scale=1.0">
  <link rel="stylesheet" href="css.css">
  <title>1. Gem Pond</title>
</head>
<body>
  <h1>Gem Pond</h1>
  <script src="main.js"></script>
  <button id="gemPond_btn" onclick="findGems()">GET GEM</button>
</body>
</html>

When I click the "GET GEM" button several times in a row, I always get the same result instead of getting a random one.

I'm not sure what I'm doing wrong here.

Thanks in advance.

What about turning randomGemStone into a function? randomGemStone = () =>? then call that in findGems? console.log("You found a " + randomGemStone() + "!"); — evolutionxbox
– evolutionxbox, Commented Apr 19, 2021 at 20:16

CherryDT · Accepted Answer · 2021-04-19 20:20:03Z

3

Move the let randomGemStone line into the findGems function:


function findGems()
{
  let randomGemStone = gemStones[Math.floor(Math.random()*gemStones.length)];
  console.log("You found a " + randomGemStone + "!");
}

Otherwise you run it only once on page load and not every time you click the button.

answered Apr 19, 2021 at 20:20

CherryDT

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2 Comments

the_previ Over a year ago

If randomGemStone value will not change you can use const instead of let

QuimB Over a year ago

Thank you. It made perfect sense once I read your comment.

Dan Wilstrop · Accepted Answer · 2021-04-19 20:24:51Z

1

 let gemStones = [
  "Amethyst",
  "Diamond",
  "Emerald",
  "Ruby",
  "Sapphire",
  "Topaz",
  "Onyx",
];


const findGems = () => {
  let randomGemStone = gemStones[Math.floor(Math.random()*gemStones.length)];
  console.log(`You found a ${randomGemStone}!`);
}

Note I have moved randomGemStone inside the function. Or the value will only be updated once when the script loads, this way it will be random everytime findGems() is called

answered Apr 19, 2021 at 20:24

Dan Wilstrop

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