Check if element is visible on screen [duplicate]

Question

Possible Duplicate:
jQuery - Check if element is visible after scroling

I'm trying to determine if an element is visible on screen. In order to to this, I'm trying to find the element's vertical position using offsetTop, but the value returned is not correct. In this case, the element is not visible unless you scroll down. But despite of this, offsetTop returns a value of 618 when my screen height is 703, so according to offsetTop the element should be visible.

The code I'm using looks like this:

function posY(obj) {   var curtop = 0;    if( obj.offsetParent )   {     while(1)     {       curtop += obj.offsetTop;        if( !obj.offsetParent )       {         break;       }        obj = obj.offsetParent;     }   } else if( obj.y )     {      curtop += obj.y;     }    return curtop; } 

Thank you in advance!

Answer 1

--- Shameless plug ---
I have added this function to a library I created vanillajs-browser-helpers: https://github.com/Tokimon/vanillajs-browser-helpers/blob/master/inView.js
-------------------------------

Well BenM stated, you need to detect the height of the viewport + the scroll position to match up with your top position. The function you are using is ok and does the job, though its a bit more complex than it needs to be.

If you don't use jQuery then the script would be something like this:

function posY(elm) {     var test = elm, top = 0;      while(!!test && test.tagName.toLowerCase() !== "body") {         top += test.offsetTop;         test = test.offsetParent;     }      return top; }  function viewPortHeight() {     var de = document.documentElement;      if(!!window.innerWidth)     { return window.innerHeight; }     else if( de && !isNaN(de.clientHeight) )     { return de.clientHeight; }          return 0; }  function scrollY() {     if( window.pageYOffset ) { return window.pageYOffset; }     return Math.max(document.documentElement.scrollTop, document.body.scrollTop); }  function checkvisible( elm ) {     var vpH = viewPortHeight(), // Viewport Height         st = scrollY(), // Scroll Top         y = posY(elm);          return (y > (vpH + st)); } 

Using jQuery is a lot easier:

function checkVisible( elm, evalType ) {     evalType = evalType || "visible";      var vpH = $(window).height(), // Viewport Height         st = $(window).scrollTop(), // Scroll Top         y = $(elm).offset().top,         elementHeight = $(elm).height();      if (evalType === "visible") return ((y < (vpH + st)) && (y > (st - elementHeight)));     if (evalType === "above") return ((y < (vpH + st))); } 

This even offers a second parameter. With "visible" (or no second parameter) it strictly checks whether an element is on screen. If it is set to "above" it will return true when the element in question is on or above the screen.

See in action: http://jsfiddle.net/RJX5N/2/

I hope this answers your question.

-- IMPROVED VERSION--

This is a lot shorter and should do it as well:

function checkVisible(elm) {   var rect = elm.getBoundingClientRect();   var viewHeight = Math.max(document.documentElement.clientHeight, window.innerHeight);   return !(rect.bottom < 0 || rect.top - viewHeight >= 0); } 

with a fiddle to prove it: http://jsfiddle.net/t2L274ty/1/

And a version with threshold and mode included:

function checkVisible(elm, threshold, mode) {   threshold = threshold || 0;   mode = mode || 'visible';    var rect = elm.getBoundingClientRect();   var viewHeight = Math.max(document.documentElement.clientHeight, window.innerHeight);   var above = rect.bottom - threshold < 0;   var below = rect.top - viewHeight + threshold >= 0;    return mode === 'above' ? above : (mode === 'below' ? below : !above && !below); } 

and with a fiddle to prove it: http://jsfiddle.net/t2L274ty/2/