Dynamically arranging divs using jQuery

Asked 2009-05-30T10:46:15.270

I have the following structure:

<div id="container">
<div id="someid1" style="float:right"></div>
<div id="someid2" style="float:right"></div>
<div id="someid3" style="float:right"></div>
<div id="someid4" style="float:right"></div>
</div>

Now someid is acually a unique id for that div. Now i receive an array which has a different order say someid 3,2,1,4, then how do i move these divs around to match the new order using jQuery?

Thankyou very much for your time.

javascript jquery

Edit Report

2 Answers

My plugin version - Working Demo

Takes an array and optional id prefix and reorders elements whose ids correspond to the order of (id prefix) + values inside the array. Any values in the array that don't have an element with the corresponding id will be ignored, and any child elements that do not have an id within the array will be removed.

(function($) {

$.fn.reOrder = function(array, prefix) {
  return this.each(function() {
    prefix = prefix || "";

    if (array) {    
      for(var i=0; i < array.length; i++) 
        array[i] = $('#' + prefix + array[i]);

      $(this).empty();  

      for(var i=0; i < array.length; i++)
        $(this).append(array[i]);      
    }
  });    
}
})(jQuery);

Code from the demo

jQuery

 $(function() {
  $('#go').click(function() {

    var order = $('#order').val() == ""? null: $('#order').val().split(",");
    $('#container').reOrder(order, 'someid');
  });
});

(function($) {

$.fn.reOrder = function(array, prefix) {
  return this.each(function() {
    prefix = prefix || "";

    if (array) {    
      for(var i=0; i < array.length; i++) 
        array[i] = $('#' + prefix + array[i]);

      $(this).empty();  

      for(var i=0; i < array.length; i++)
        $(this).append(array[i]);      
    }
  });    
}
})(jQuery);

HTML

<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN"
  "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml" xml:lang="en" lang="en">
<head>
<script src="http://ajax.googleapis.com/ajax/libs/jquery/1.3.2/jquery.min.js"></script>
<title>reOrder Demo</title>
<meta http-equiv="Content-type" content="text/html; charset=utf-8" />
<style type="text/css" media="screen">
body { backgr


                        
                        
                            answered 2009-05-30T11:51:00.583


            
            
                
                    
                        
                            
                            
                        
                        3
                        
                            
                            
                        
                        
                    
                    
                        
                        [Edit], This is tested and works:

var order = [3,2,1,4];
var container = $("#container");
var children = container.children();
container.empty();
for (var i = 0; i < order.length; i++){
    container.append(children[order[i]-1])
}


The i-1 is necessary since your ordering starts at 1 but arrays are indexed from 0.

Thanks to J-P and Russ Cam for making me look at it again.

                        
                        
                            answered 2009-05-30T11:00:52.820



        
        
            Your Answer