I would like to get the one element which is the most visible on the screen (takes up the most space). I have added an example picture below to understand my question a bit
Yes, this question is too broad. But I was interested on solving it. Here is crude example on how to accomplish it.
I tried to explain what's going on with comments. It surely can be done better, but I hope it helps.
// init on page ready
$(function() {
// check on each scroll event
$(window).scroll(function(){
// elements to be tested
var _elements = $('.ele');
// get most visible element (result)
var ele = findMostVisible(_elements);
});
});
function findMostVisible(_elements) {
// find window top and bottom position.
var wtop = $(window).scrollTop();
var wbottom = wtop + $(window).height();
var max = 0; // use to store value for testing
var maxEle = false; // use to store most visible element
// find percentage visible of each element
_elements.each(function(){
// get top and bottom position of the current element
var top = $(this).offset().top;
var bottom = top + $(this).height();
// get percentage of the current element
var cur = eleVisible(top, bottom, wtop, wbottom);
// if current element is more visible than previous, change maxEle and test value, max
if(cur > max) {
max = cur;
maxEle = $(this);
}
});
return maxEle;
}
// find visible percentage
function eleVisible(top, bottom, wtop, wbottom) {
var wheight = wbottom - wtop;
// both bottom and top is vissible, so 100%
if(top > wtop && top < wbottom && bottom > wtop && bottom < wbottom)
{
return 100;
}
// only top is visible
if(top > wtop && top < wbottom)
{
return 100 + (wtop - top) / wheight * 100;
}
// only bottom is visible
if(bottom > wtop && bottom < wbottom)
{
return 100 + (bottom - wbottom) / wheight * 100;
}
// element is not visible
return 0;
}
Working example - https://jsfiddle.net/exabyssus/6o30sL24/