How to check if an element is off-screen

Question

I need to check with jQuery if a DIV element is not falling off-screen. The elements are visible and displayed according CSS attributes, but they could be intentionally placed off-screen by:

position: absolute; 
left: -1000px; 
top: -1000px;

I could not use the jQuery :visible selector as the element has a non-zero height and width.

I am not doing anything fancy. This absolute position placement is the way my Ajax framework implements the hide/show of some widgets.

Answer 1

Depends on what your definition of "offscreen" is. Is that within the viewport, or within the defined boundaries of your page?

Using Element.getBoundingClientRect() you can easily detect whether or not your element is within the boundries of your viewport (i.e. onscreen or offscreen):

jQuery.expr.filters.offscreen = function(el) {
  var rect = el.getBoundingClientRect();
  return (
           (rect.x + rect.width) < 0 
             || (rect.y + rect.height) < 0
             || (rect.x > window.innerWidth || rect.y > window.innerHeight)
         );
};

You could then use that in several ways:

// returns all elements that are offscreen
$(':offscreen');

// boolean returned if element is offscreen
$('div').is(':offscreen');

Answer 2

There's a jQuery plugin here which allows users to test whether an element falls within the visible viewport of the browser, taking the browsers scroll position into account.

$('#element').visible();

You can also check for partial visibility:

$('#element').visible( true);

One drawback is that it only works with vertical positioning / scrolling, although it should be easy enough to add horizontal positioning into the mix.

Answer 3

No need for a plugin to check if outside of view port.

var w = Math.max(document.documentElement.clientWidth, window.innerWidth || 0)
var h = Math.max(document.documentElement.clientHeight, window.innerHeight || 0)
var d = $(document).scrollTop();

$.each($("div"),function(){
    p = $(this).position();
    //vertical
    if (p.top > h + d || p.top > h - d){
        console.log($(this))
    }
    //horizontal
    if (p.left < 0 - $(this).width() || p.left > w){
        console.log($(this))
    }
});

Answer 4

Well... I've found some issues in every proposed solution here.

• You should be able to choose if you want entire element to be on screen or just any part of it
• Proposed solutions fails if element is higher/wider than window and kinda covers browser window.

Here is my solution that include jQuery .fn instance function and expression. I've created more variables inside my function than I could, but for complex logical problem I like to divide it into smaller, clearly named pieces.

I'm using getBoundingClientRect method that returns element position relatively to the viewport so I don't need to care about scroll position

Useage:

$(".some-element").filter(":onscreen").doSomething();
$(".some-element").filter(":entireonscreen").doSomething();
$(".some-element").isOnScreen(); // true / false
$(".some-element").isOnScreen(true); // true / false (partially on screen)
$(".some-element").is(":onscreen"); // true / false (partially on screen)
$(".some-element").is(":entireonscreen"); // true / false 

Source:

$.fn.isOnScreen = function(partial){

    //let's be sure we're checking only one element (in case function is called on set)
    var t = $(this).first();

    //we're using getBoundingClientRect to get position of element relative to viewport
    //so we dont need to care about scroll position
    var box = t[0].getBoundingClientRect();

    //let's save window size
    var win = {
        h : $(window).height(),
        w : $(window).width()
    };

    //now we check against edges of element

    //firstly we check one axis
    //for example we check if left edge of element is between left and right edge of scree (still might be above/below)
    var topEdgeInRange = box.top >= 0 && box.top <= win.h;
    var bottomEdgeInRange = box.bottom >= 0 && box.bottom <= win.h;

    var leftEdgeInRange = box.left >= 0 && box.left <= win.w;
    var rightEdgeInRange = box.right >= 0 && box.right <= win.w;


    //here we check if element is bigger then window and 'covers' the screen in given axis
    var coverScreenHorizontally = box.left <= 0 && box.right >= win.w;
    var coverScreenVertically = box.top <= 0 && box.bottom >= win.h;

    //now we check 2nd axis
    var topEdgeInScreen = topEdgeInRange && ( leftEdgeInRange || rightEdgeInRange || coverScreenHorizontally );
    var bottomEdgeInScreen = bottomEdgeInRange && ( leftEdgeInRange || rightEdgeInRange || coverScreenHorizontally );

    var leftEdgeInScreen = leftEdgeInRange && ( topEdgeInRange || bottomEdgeInRange || coverScreenVertically );
    var rightEdgeInScreen = rightEdgeInRange && ( topEdgeInRange || bottomEdgeInRange || coverScreenVertically );

    //now knowing presence of each edge on screen, we check if element is partially or entirely present on screen
    var isPartiallyOnScreen = topEdgeInScreen || bottomEdgeInScreen || leftEdgeInScreen || rightEdgeInScreen;
    var isEntirelyOnScreen = topEdgeInScreen && bottomEdgeInScreen && leftEdgeInScreen && rightEdgeInScreen;

    return partial ? isPartiallyOnScreen : isEntirelyOnScreen;

};

$.expr.filters.onscreen = function(elem) {
  return $(elem).isOnScreen(true);
};

$.expr.filters.entireonscreen = function(elem) {
  return $(elem).isOnScreen(true);
};

Answer 5

I know this is kind of late but this plugin should work. http://remysharp.com/2009/01/26/element-in-view-event-plugin/

$('p.inview').bind('inview', function (event, visible) {
if (visible) {
  $(this).text('You can see me!');
} else {
  $(this).text('Hidden again');
}

Answer 6

• Get the distance from the top of the given element
• Add the height of the same given element. This will tell you the total number from the top of the screen to the end of the given element.

• Then all you have to do is subtract that from total document height

  jQuery(function () {
      var documentHeight = jQuery(document).height();
      var element = jQuery('#you-element');
      var distanceFromBottom = documentHeight - (element.position().top + element.outerHeight(true));
      alert(distanceFromBottom)
  });
  

Answer 7

You could check the position of the div using $(div).position() and check if the left and top margin properties are less than 0 :

if($(div).position().left < 0 && $(div).position().top < 0){
    alert("off screen");
}