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问题:

I am attempting to display a series of images in a random order. However, I do not want any single item to repeat until all items have been shown, so instead of selecting a random image from the array, I want to take the entire array, randomize it, and then select in sequence from the first to the last element. Here's my code:

HTML:

<div id="tout4"
<img src="images/gallery01.jpg" class="img_lg"/>
<img src="images/gallery02.jpg" class="img_lg"/>
<img src="images/gallery03.jpg" class="img_lg"/>
</div>

and the javascript, which currently selects and displays the items in order:

var galleryLength = $('#tout4 img.img_lg').length;
var currentGallery = 0;
setInterval(cycleGallery, 5000);


function cycleGallery(){

    $('#tout4 img.img_lg').eq(currentGallery).fadeOut(300);

    if (currentGallery < (galleryLength-1)){
        currentGallery++;
    } else {
        currentGallery = 0;
    }

    $('#tout4 img.img_lg').eq(currentGallery).fadeIn(300);
}

So how do I rearrange the actual order of the images, and not just the order in which they are selected?

回答1:

After much exploration, I decided to take the fisher-yates algorithm and apply it with jquery without requiring cloning, etc.

$('#tout4 img.img_lg').shuffle();

/*
* Shuffle jQuery array of elements - see Fisher-Yates algorithm
*/
jQuery.fn.shuffle = function () {
    var j;
    for (var i = 0; i < this.length; i++) {
        j = Math.floor(Math.random() * this.length);
        $(this[i]).before($(this[j]));
    }
    return this;
};

回答2:

You can also use the common JavaScript Array randomize sorter, also commented here and here:

$('<my selector>').sort( function(){ return ( Math.round( Math.random() ) - 0.5 ) } );

回答3:

Ended up using this (thanks Blair!) -

/**
 * jQuery Shuffle (/web/20120307220753/http://mktgdept.com/jquery-shuffle)
 * A jQuery plugin for shuffling a set of elements
 *
 * v0.0.1 - 13 November 2009
 *
 * Copyright (c) 2009 Chad Smith (/web/20120307220753/http://twitter.com/chadsmith)
 * Dual licensed under the MIT and GPL licenses.
 * /web/20120307220753/http://www.opensource.org/licenses/mit-license.php
 * /web/20120307220753/http://www.opensource.org/licenses/gpl-license.php
 *
 * Shuffle elements using: $(selector).shuffle() or $.shuffle(selector)
 *
 **/
(function(d){d.fn.shuffle=function(c){c=[];return this.each(function(){c.push(d(this).clone(true))}).each(function(a,b){d(b).replaceWith(c[a=Math.floor(Math.random()*c.length)]);c.splice(a,1)})};d.shuffle=function(a){return d(a).shuffle()}})(jQuery);

So then the only additions that need to be made to the above code are to include the script, and call the shuffle function:

<script type="text/javascript" src="js/jquery-shuffle.js"></script>
$('#tout4 img.img_lg').shuffle();

回答4:

<!DOCTYPE html>
<html>
<body>

<p id="demo">Click the button to sort the array.</p>

<button onclick="myFunction()">Try it</button>

<script>
  function myFunction()
  {
    var points = [40,100,1,5,25,10];
    points.sort(function(a,b){return (Math.random()-0.5)});
    var x=document.getElementById("demo");
    x.innerHTML=points;
  }
</script>

</body>
</html>

Explanation: normally you have "return (a-b)" yielding a positive number for ascending sort order; or you have "return (b-a)" yielding a negative number for descending sort order.

Here we use Math.random()-0.5 which gives in half of the cases a positive number and in half of the cases a negative number. Thus sorting of pairs is either ascending or descending, yielding a random distribution of the array elements.