How to print 52 randomly generated numbers to the console?

Question

Simple code, working on a card game, however I am encountering this error right at the beginning. I am trying to generate 52 random numbers and print out the appropriate array element associated with it. Here is my code:

    <!DOCTYPE html>
    <html>
    <head>

    </head>
    <body>
        <button onclick="shuffleCards()">Shuffle Cards</button>
        <script>
            function shuffleCards() {
                var cardDeck = ["1S", "1C", "1H", "1D", "2S", "2C", "2H", "2D", "3S", "3C", "3H", "3D", "4S", "4C", "4H", "4D", "5S", "5C", "5H", "5D", "6S", "6C", "6H", "6D", "7S", "7C", "7H", "7D", "8S", "8C", "8H", "8D", "9S", "9C", "9H", "9D", "10S", "10C", "10H", "10D", "11S", "11C", "11H", "11D", "12S", "12C", "12H", "12D", "13S", "13C", "13H", "13D", ];
                var handUser = [];
                var handCPU = [];
                var shuffledDeck = [];
                for (var i = 51; i < 1; i--) {
                    var randomNumber = Math.floor(Math.random()*(i + 1));
                    console.log(cardDeck[randomNumber]);
                }
            }
        </script>
    </body>
</html>

In your for loop, `for (var i = 51; i < 1; i--)` i is never less than 1. Maybe it should be >1 — jgabb, Mar 29 '16 at 21:16
Also, if you want each card appear only once, take a look at http://stackoverflow.com/questions/6274339/how-can-i-shuffle-an-array-in-javascript — Aleph Aleph, Mar 29 '16 at 21:18

score 0 · Answer 1 · answered Mar 29 '16 at 21:19

0

The for loop iterator, i, starts out greater than 1, and so it isn't less than 1. Change your operator to > instead of <

answered Mar 29 '16 at 21:19

David Kluszczynski

39
1
11

score 0 · Answer 2 · answered Mar 29 '16 at 21:20

0

It's quite simple really, your for loop's condition is never true, it should be i >== 0. If i = 51 then i < 1 will never be true.

answered Mar 29 '16 at 21:20

AGE

3,752
3
38
60

score 0 · Accepted Answer · answered Mar 29 '16 at 21:20

A problem with your solution is that it doesn't guarantee that a unique card will be returned each time, which I assume you want to guarantee.

If this is true, then this solution will iterate 52 times, each time taking and removing a card from the deck using the splice method:

var cardDeck = ["1S", "1C", "1H", "1D", "2S", "2C", "2H", "2D", "3S", "3C", "3H", "3D", "4S", "4C", "4H", "4D", "5S", "5C", "5H", "5D", "6S", "6C", "6H", "6D", "7S", "7C", "7H", "7D", "8S", "8C", "8H", "8D", "9S", "9C", "9H", "9D", "10S", "10C", "10H", "10D", "11S", "11C", "11H", "11D", "12S", "12C", "12H", "12D", "13S", "13C", "13H", "13D", ];
var total = cardDeck.length;
for (var i = 0; i < total; i++) {
  var randomPos = Math.floor((Math.random() * (cardDeck.length-1)));
  var randomCard = cardDeck.splice(randomPos, 1).pop();
  console.log(randomCard);
}

Thanks for the quick response, I know it will not yet return a unique card. I am just trying to get this loop to function first before I splice out elements from the array with each iteration. Thanks for your help. — JSON_Derulo, Mar 29 '16 at 21:35

score 0 · Answer 4 · answered Mar 29 '16 at 21:20

0

<!DOCTYPE html>
<html>
<head>
<script>
function shuffleCards() {
  console.log('test')
  var cardDeck = ["1S", "1C", "1H", "1D", "2S", "2C", "2H", "2D", "3S",
  "3C", "3H", "3D", "4S", "4C", "4H", "4D", "5S", "5C", "5H", "5D", "6S", "6C",
  "6H", "6D", "7S", "7C", "7H", "7D", "8S", "8C", "8H", "8D", "9S", "9C", "9H", "9D",
  "10S", "10C", "10H", "10D", "11S", "11C", "11H", "11D", "12S", "12C", "12H", "12D",
  "13S", "13C", "13H", "13D" ];
  var handUser = [];
  var handCPU = [];
  var shuffledDeck = [];
  for (var i = 51; i > 0; i--) {
    var randomNumber = Math.floor(Math.random()*(i + 1));
    console.log(cardDeck[randomNumber]);
  }
}
</script>
</head>
<body>
  <button onclick="shuffleCards()">Shuffle Cards</button>
</body>
</html>

This works...

answered Mar 29 '16 at 21:20

Arash Saidi

2,228
20
36

@Will Where are you running it? – Arash Saidi Mar 29 '16 at 23:04
Your logic doesn't work. The loop does not give 52 random cards. – Will Mar 29 '16 at 23:46
@Will The question wasn't 52 random cards, but 52 random numbers that return cards from a list. Semantics I know, but his issue was with the for loop. – Arash Saidi Mar 30 '16 at 06:58

score 0 · Answer 5 · answered Mar 29 '16 at 23:05

Here's one more method for shuffling the cards. This method shows an approach that does not splice the original array. For an small array, it's quick and easy to clone the array and remove items from it as necessary, but if we were dealing with a really big array, when using an object to track used items is a good approach.

function shuffleCards() {
    var cardDeck = ["1S", "1C", "1H", "1D", "2S", "2C", "2H", "2D", "3S", "3C", "3H", "3D",
            "4S", "4C", "4H", "4D", "5S", "5C", "5H", "5D", "6S", "6C", "6H", "6D",
            "7S", "7C", "7H", "7D", "8S", "8C", "8H", "8D", "9S", "9C", "9H", "9D",
            "10S", "10C", "10H", "10D", "11S", "11C", "11H", "11D",
            "12S", "12C", "12H", "12D", "13S", "13C", "13H", "13D"];
    var used = {};
    var shuffledDeck = [];
    while (shuffledDeck.length < 52) {
        var randomNumber = Math.floor(Math.random() * (51 + 1));
        if (!used[randomNumber]) {
            shuffledDeck.push(cardDeck[randomNumber]);
            used[randomNumber] = true;
        }
    }
    console.log(shuffledDeck.length, shuffledDeck);
}
shuffleCards();

How to print 52 randomly generated numbers to the console?

5 Answers5