Sort random field items via jQuery

Question

I have a HTML slide menu. With the following:

<div class="slide">
<img typeof="foaf:Image" class="image-style-mylogo" src="http://site.com/1.png" alt="">
<img typeof="foaf:Image" class="image-style-mylogo" src="http://site.com/2.png" alt="">
<img typeof="foaf:Image" class="image-style-mylogo" src="http://site.com/3.png" alt="">
<img typeof="foaf:Image" class="image-style-mylogo" src="http://site.com/4.png" alt="">
<img typeof="foaf:Image" class="image-style-mylogo" src="http://site.com/5.png" alt="">
<img typeof="foaf:Image" class="image-style-mylogo" src="http://site.com/6.png" alt="">
</div>

And i want, get images with random sort every refresh. I used this code:

function reorder() {
    var grp = $(".slide").children();
    var cnt = grp.length;

    var temp, x;
    for (var i = 0; i < cnt; i++) {
        temp = grp[i];
        x = Math.floor(Math.random() * cnt);
        grp[i] = grp[x];
        grp[x] = temp;
    }
    $(grp).remove();
    $(".slide").append($(grp));
}

function orderPosts() {
    $(".slide").html(orig);
}

But don't work. What might I be doing wrong?

How do you call these functions? And btw, no need for `remove`, just `$(".slide").append( grp );` is enough — Bergi, Dec 21 '12 at 13:53

score 2 · Answer 1 · answered Dec 21 '12 at 13:58

2

There is a much simpler way to do this. Since a jQuery collection is Array-like, you can call native Array prototypes on them. So using the native Array.sort, your code you be rewritten like this:

var grp = $(".slide").children(); // the original collection, if you want to save it...

Array.prototype.sort.call(grp, function() {
    return Math.round(Math.random())-0.5;
});

$('.slide').empty().append(grp);

Here is a demo: http://jsfiddle.net/JTGfC/

answered Dec 21 '12 at 13:58

David Hellsing

106,495
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Why don't you use `grp.sort(sortFunction);`? – Cerbrus Dec 21 '12 at 14:04
@Cebrus I don’t think `jQuery.fn.sort` exists in the API – David Hellsing Dec 21 '12 at 14:06
@Cerbrus Although it does work, and I can see that it’s implemented "for internal use": https://github.com/jquery/jquery/blob/master/src/core.js#L280 – David Hellsing Dec 21 '12 at 14:09

score 1 · Answer 2 · answered Dec 21 '12 at 13:58

I'm not sure whether this is 100% kosher, but you could apply Fisher-Yates here, without dependency on jQuery.

fisherYates(document.getElementsByClassName('slide')[0]);

// Fisher-Yates, modified to shuffle DOM container's children instead of an array.
function fisherYates (node)
{
  var children = node.children,
  n = children.length;

  if (!n) {
    return false;
  }

  while (--n) {
     var i = Math.floor( Math.random() * ( n + 1 ) ),
     c1 = children[n];

     node.insertBefore(c1, children[i]);
   }
}

Demo

score 0 · Accepted Answer · answered Dec 21 '12 at 13:56

0

Your reorder() function is fine, I tested it in this fiddle : http://jsfiddle.net/kokoklems/YpjwE/

I did not used the second function orderPosts() though...

answered Dec 21 '12 at 13:56

clement g

461
1
3
10

score 0 · Answer 4 · edited May 23 '17 at 11:56

This should do it for you:

function reorder() {
    var grp = $(".slide").children();
    var cnt = grp.length;

    var indexes = [];
    // Build a array from 0 to cnt-1 
    for (var i = 0; i < cnt; i++) {
        indexes.push(i);
    } 
    // Shuffle this array. This random array of indexes will determine in what order we add the images.
    indexes = shuffle(indexes);

    // Add the images. (No need to remove them first, .append just moves them)
    for (var i = 0; i < cnt; i++) {
        $(".slide").append($(grp[indexes[i]]));
    }
}

function shuffle(o){
    for(var j, x, i = o.length; i; j = parseInt(Math.random() * i), x = o[--i], o[i] = o[j], o[j] = x);
    return o;
}

Shuffle function source

Working sample (I used spans instead of the images, to show the result better)

Sort random field items via jQuery

4 Answers4

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