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Home/ Questions/Q 9145209
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Editorial Team
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Editorial Team
Asked: June 17, 20262026-06-17T10:27:27+00:00 2026-06-17T10:27:27+00:00

I’m looking for a very fast solution to a div scrolling problem. I have

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I’m looking for a very fast solution to a div scrolling problem.

I have a set of divs, like forum posts, that are laid out one on top of the other. As the page scrolls down or up, I’d like to know when one of those divs hit’s an arbitrary point on the page.

One way I tried was adding an onScroll event to each item, but as the number of items grow the page really starts to lag.

Anyone know a more efficient way to do this? Thanks /w

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-06-17T10:27:28+00:00Added an answer on June 17, 2026 at 10:27 am

    Well, I’m new to all this, so may be someone should correct me 🙂

    I propose to

    • cache posts position
    • caсhe current
    • use binary search

    Demo: http://jsfiddle.net/zYe8M/

    <div class="post"></div>
    <div class="post"></div>
    <div class="post"></div>
    

    …

    var posts = $(".post"), // our elements
        postsPos = [], // caсhe for positions
        postsCur = -1, // cache for current
        targetOffset = 50; // position from top of window where you want to make post current
    
    // filling postsPos with positions
    posts.each(function(){
        postsPos.push($(this).offset().top);
    });
    
    // on window scroll
    $(window).bind("scroll", function(){
      // get target post number
      var targ = postsPos.binarySearch($(window).scrollTop() + targetOffset);
      // only if we scrolled to another post
      if (targ != postsCur) {
        // set new cur
        postsCur = targ;
        // moving cur class
        posts.removeClass("cur").eq(targ).addClass("cur");
      }
    });
    
    // binary search with little tuning on return to get nearest from bottom
    Array.prototype.binarySearch = function(find) {
      var low = 0, high = this.length - 1,
          i, comparison;
      while (low <= high) {
        i = Math.floor((low + high) / 2);
        if (this[i] < find) { low = i + 1; continue; };
        if (this[i] > find) { high = i - 1; continue; };
        return i;
      }
      return this[i] > find ? i-1 : i;
    };
    
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