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Home/ Questions/Q 6201115
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Editorial Team
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Editorial Team
Asked: May 24, 20262026-05-24T04:29:43+00:00 2026-05-24T04:29:43+00:00

My problem is this. I have a DIV which replicates an image using CSS3.

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My problem is this. I have a DIV which replicates an image using CSS3. This Div needs to be repeated at a set interval down one side of the page. I.e. every 200px. Is there a way to for this to be automated. Normally I would use a repeating bgimage but in this case no images are allowed. I also could repeat the div each time but it’s a large page that will be updated frequently with more content.

Any ideas would be very much appreciated.

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  1. Editorial Team
    Editorial Team
    2026-05-24T04:29:43+00:00Added an answer on May 24, 2026 at 4:29 am

    I would do it like this:

    document height / element height + 200px = x
    place the element X times in to you document.

    That’s it.

    Spoken in jQuery this looks like that:

    try it out here: http://jsfiddle.net/aF68z/

    var repeatMe = function ( $o, space ){
      var oHeight,dHeight, multiplicator, res, html, $parent;
    
      dHeight = $(document).height(); //height of you document
      oHeight = $o.height(); //height of the element that shoud repeat 
      multiplicator = Math.floor(dHeight / (oHeight + space)); //how many time the element can repeat (including the margin)
      $parent = $o.parent(); //gets the parent that finally will hold all repeating items
      html = $parent.html(); //gets the HTML code of the element that repeats
    
      /* appending and cloning are very CPU heavy and it makes no sense to do so only for a visual matter, "string" + "string" etc... is very slow if the string becomes long, this is a simple trick how to avoid this: */
      res = [];  
      for (var i = 0; i < multiplicator; ++ i) {
        res.push(html);
      }
      html = res.join("");
    
    
      $parent.html(html); //appending the HTML of the all the repeated elements to the parent again.
    };
    
    repeatMe( $("div.deco div:eq(0)"),200 );
    

    PS: At least try something on your own next time.

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