Any way to shuffle content in multiple div elements

Question

I'm relatively new to Javascript and was wondering if there's a quick way to shuffle content that is contained in multiple <div> tags. For example

<div id='d1'>
  <span>alpha</span>
  <img src='alpha.jpg'>
</div>
<div id='d2'>
  <span>beta</span>
  <img src='beta.jpg'>
</div>
<div id='d3'>
  <span>gamma</span>
  <img src='gamma.jpg'>
</div>

<button onclick='shuffle_content();'>Shuffle</button>

After clicking on the button, I'd like the content in d1, d2, d3 to change places (for example maybe d3 would be first, then d1, then d2).

A quick way to kind of move things around is to copy the first div element (d1), then put it at the very end (after d3), and then delete the original d1. But that doesn't really randomize things. It just makes things go in the cycle (which might be ok).

Any suggestions would be appreciated. Thanks.

Seclude

Answer 1

are you ok with using a javascript library like jQuery? here's a quick jQuery example to accomplish what you're after. the only modification to your HTML is the addition of a container element as suggested:

<div id="shuffle">
    <div id='d1'>...</div>
    <div id='d2'>...</div>
    <div id='d3'>...</div>
</div>

and javascript:

function shuffle(e) {               // pass the divs to the function
    var replace = $('<div>');
    var size = e.size();

    while (size >= 1) {
       var rand = Math.floor(Math.random() * size);
       var temp = e.get(rand);      // grab a random div from our set
       replace.append(temp);        // add the selected div to our new set
       e = e.not(temp); // remove our selected div from the main set
       size--;
    }
    $('#shuffle').html(replace.html() );     // update our container div with the
                                             // new, randomized divs
}

shuffle( $('#shuffle div') );

Hoist

Thanks Owen, I think this is best answer I've seen so far. – Seclude

Unfortunately, since my reputation is below 15, I can't vote this up. But if I could, I would. – Seclude

Answer 2

A recent question was just closed as duplicate of this, but I feel I've got a better answer than any here. This method is very direct. There's no mucking with copying HTML, thus preserving changes to the DOM, styles, event handlers, etc.

To shuffle all the children of some parent element, select a random child and append it back to the parent one at a time until all the children have been re-appended.

Using jQuery:

var parent = $("#shuffle");
var divs = parent.children();
while (divs.length) {
    parent.append(divs.splice(Math.floor(Math.random() * divs.length), 1)[0]);
}

Demo: http://jsfiddle.net/C6LPY/2

Without jQuery it's similar and just as simple:

var parent = document.getElementById("shuffle");
var divs = parent.children;
var frag = document.createDocumentFragment();
while (divs.length) {
    frag.appendChild(divs[Math.floor(Math.random() * divs.length)]);
}
parent.appendChild(frag);

Demo: http://jsfiddle.net/C6LPY/5/

---

Edit: Here's a break down of the code:

// Create a document fragment to hold the shuffled elements
var frag = document.createDocumentFragment();

// Loop until every element is moved out of the parent and into the document fragment
while (divs.length) {

    // select one random child element and move it into the document fragment
    frag.appendChild(divs[Math.floor(Math.random() * divs.length)]);
}

// appending the document fragment appends all the elements, in the shuffled order
parent.appendChild(frag);

Allseed

Good answer. If Stackoverflow allowed questions to have multiple "accepted" answers, I would also label this as "accepted". – Seclude

is this also applicable for shuffling list? – Kneeland

@Allseed A question on "divs.splice(Math.floor(Math.random() * divs.length), 1)[0]" what is the use of [0] ? – Kneeland

@Kneeland - .splice() returns an array containing, in this case, a single element. [0] gets you the first element in that array. – Allseed

@Kneeland - yes, you can use this to shuffle a list of any kind of element. – Allseed

But splice removes the content right? and returns the removed items? – Kneeland

@Kneeland - Yes, that's the point. Notice the while loop... we loop until the Array is empty and all the elements have been re-inserted. – Allseed

@Kneeland - I've updated my answer with a breakdown of the exactly what's happening inside the while loop. I hope that helps to clarify it for you. – Allseed

when will the loop actually stop? aren't the loop going to stop since there would be always one entry or one item in the divs array? – Kneeland

@Kneeland - No, remember .splice() removes the item from the array. Since in each iteration of the while loop we remove exactly one item from the array, the loop will end as soon as we have processed each element. divs is an Array whose elements reference dom nodes. We created that array by copying the elements from the parent element's children collection. Removing an element from the array does not modify the DOM. Ie, calling divs.splice(n, 1) removes an element from the Array, but the element still exists in the page as a child of parent. – Allseed

Answer 3

You can grab the content of each div

c1 = document.getElementById('div1').innerHTML
c2 = document.getElementById('div2').innerHTML
c3 = document.getElementById('div3').innerHTML

Then determine a new order for them randomly .. and then put each content in the new destination

say for instance, the randomness gave:

c1_div = 'div2'
c2_div = 'div1'
c3_div = 'div3'

then you just:

document.getElementById(c1_div).innerHTML = c1
document.getElementById(c2_div).innerHTML = c2
document.getElementById(c3_div).innerHTML = c3

Aprilette

Answer 4

Expanding on the nice answer by @gilly3, using jQuery one can actually avoid appending randomly-chosen elements of divs in a loop, by randomly sorting divinstead and appending them all at once:

$(function() {
  var parent = $("#shuffle");
  var divs = parent.children();
  divs.sort(function(a, b) {
    return 0.5 - Math.random();
  });
  parent.append(divs);
});

Demo: http://jsfiddle.net/ey70Lxhk/

Note however that this technique is not accurate in terms of randomness, and relies on sort which does not scale linearly with the number of elements.

Lenssen

Answer 5

I'd wrap the divs in an outer div, then pass its id to shuffle_content().

In there, you could create a new div, cloning the wrapper div's nodes in a random order to fill it, then replace the wrapper div with the new div.

Trice

Answer 6

I'd use server side code to accomplish this. I know this isn't really an answer to your question, but it is an alternative implementation.

Best Regards,
Frank

Calcutta

Answer 7

For your HTML, the short answer to your question is:

function shuffle_content() {
 var divA = new Array(3);
 for(var i=0; i < 3; i++) {
  divA[i] = document.getElementById('d'+(i+1));
  document.body.removeChild(divA[i]);
 }
 while (divA.length > 0)
  document.body.appendChild(divA.splice(Math.floor(Math.random() * divA.length),1)[0]);
}

To get there I wrote the following, which I think works better:

<html>
<div id="cards">
<div id="card0">Card0</div><div id="card1">Card1</div>
<div id="card2">Card2</div><div id="card3">Card3</div>
<div id="card4">Card4</div><div id="card5">Card5</div>
<div id="card6">Card6</div><div id="card7">Card7</div>
<div id="card8">Card8</div><div id="card9">Card9</div>
</div>
<button id="shuffle">Shuffle</button>
<script language="javascript">
<!--
document.getElementById('shuffle').onclick = function () {
var divCards = document.getElementById('cards');
var divCardsArray = new Array(
    document.getElementById('card0'),
    document.getElementById('card1'),
    document.getElementById('card2'),
    document.getElementById('card3'),
    document.getElementById('card4'),
    document.getElementById('card5'),
    document.getElementById('card6'),
    document.getElementById('card7'),
    document.getElementById('card8'),
    document.getElementById('card9')
    );
return function() {
    var mDivCardsArray=divCardsArray.slice();
    while (divCards.childNodes.length > 0) {
        divCards.removeChild(divCards.firstChild);
    }
    while (mDivCardsArray.length > 0) {
        var i = Math.floor(Math.random() * mDivCardsArray.length);
        divCards.appendChild(mDivCardsArray[i]);
        mDivCardsArray.splice(i,1);
    }
    return false;
}
}()
//-->
</script>
</html>

I was trying to pack down that last while statement to:

    while (mDivCardsArray.length > 0) {
        divCards.appendChild(
            mDivCardsArray.splice(
                Math.floor(Math.random() * mDivCardsArray.length)
                ,1)[0]
        );
    }

but this is pretty hard to read and prone to error.

Going with jQuery or Prototype you could follow the same basic structure and get the result you're looking for.

Personally, I think it looks even better if you add 2 more divs to the cards stack, expand the divCardsArray, insert the following style block, and add this code right after the divCardsArray definition.

<html>
...
<style>
html,body{height:100%;width:100%;text-align:center;font-family:sans-serif;}
#cards,#cards div{padding:5px;margin:5px auto 5px auto;width:100px;}
</style>
...
<div id="cardA">CardA</div><div id="cardB">CardB</div>
...
var colorCardsArray = new Array(
    '#f00', '#f80', '#ff0', '#8f0', '#0f0', '#0f8',
    '#0ff', '#08f', '#00f', '#80f', '#f0f', '#f08' );
for(var i=0;i<divCardsArray.length;i++)
    divCardsArray[i].style.backgroundColor=colorCardsArray[i];
...
</html>

Jannery

Answer 8

I would suggest you randomize the content, not the actual Divs themselves. You could accomplish this by putting the content in separate html pages - no header info or body, just the content.

Then use a function on page load to randomly assign which div gets what content and use this to change the DIV's content:

<script type="text/javascript">
    function ajaxManager(){
        var args = ajaxManager.arguments;

        if (document.getElementById) {
            var x = (window.ActiveXObject) ? new ActiveXObject("Microsoft.XMLHTTP") : new XMLHttpRequest();
        }
        if (x){         
            switch (args[0]){
                case "load_page":
                    if (x)
                    {
                        x.onreadystatechange = function()
                        {
                            if (x.readyState == 4 && x.status == 200){
                                el = document.getElementById(args[2]);
                                el.innerHTML = x.responseText;
                            }
                        }
                        x.open("GET", args[1], true);
                        x.send(null);
                    }
                    break;

                case "random_content":
                    ajaxManager('load_page', args[1], args[2]); /* args[1] is the content page, args[2] is the id of the div you want to populate with it. */       
                    break;
            } //END SWITCH
        } //END if(x)
    } //END AjaxManager

</script>

Define

This doesn't answer the question, just explains a different way to move the content. – Glennglenna

I offered a quick way to move the content around which is the OP's original intent. What did you read the question to ask? – Define