Check if element is visible in DOM

Is there any way that I can check if an element is visible in pure JS (no jQuery) ?

So, for example, in this page: Performance Bikes, if you hover over Deals (on the top menu), a window of deals appear, but at the beginning it was not shown. It is in the HTML but it is not visible.

So, given a DOM element, how can I check if it is visible or not? I tried: window.getComputedStyle(my_element)['display']); but it doesn't seem to be working. I wonder which attributes should I check. It comes to my mind:

display !== 'none'
visibility !== 'hidden'

Any others that I might be missing?

---

According to this MDN documentation, an element's offsetParent property will return null whenever it, or any of its parents, is hidden via the display style property. Just make sure that the element isn't fixed. A script to check this, if you have no 'position:fixed;' elements on your page, might look like:

//Where el is the DOM element you'd like to test for visibility
function isHidden(el) {
    return (el.offsetParent === null)
}

On the other hand, if you do have position fixed elements that might get caught in this search, you will sadly (and slowly) have to use window.getComputedStyle(). The function in that case might be:

//Where el is the DOM element you'd like to test for visibility
function isHidden(el) {
    var style = window.getComputedStyle(el);
    return (style.display === 'none')
}

Option #2 is probably a little more straightforward since it accounts for more edge cases, but I bet its a good deal slower, too, so if you have to repeat this operation many times, best to probably avoid it.

---

All the other solutions broke for some situation for me..

See the winning answer breaking at:

http://plnkr.co/edit/6CSCA2fe4Gqt4jCBP2wu?p=preview

Eventually, I decided that the best solution was $(elem).is(':visible') - however, this is not pure javascript. it is jquery..

so I peeked at their source and found what I wanted

jQuery.expr.filters.visible = function( elem ) {
    return !!( elem.offsetWidth || elem.offsetHeight || elem.getClientRects().length );
};

This is the source: https://github.com/jquery/jquery/blob/master/src/css/hiddenVisibleSelectors.js

---

This may help : Hide the element by positioning it on far most left position and then check the offsetLeft property. If you want to use jQuery you can simply check the :visible selector and get the visibility state of the element.

HTML :

<div id="myDiv">Hello</div>

CSS :

<!-- for javaScript-->
#myDiv{
   position:absolute;
   left : -2000px;
}

<!-- for jQuery -->
#myDiv{
    visibility:hidden;
}

javaScript :

var myStyle = document.getElementById("myDiv").offsetLeft;

if(myStyle < 0){
     alert("Div is hidden!!");
}

jQuery :

if(  $("#MyElement").is(":visible") == true )
{  
     alert("Div is hidden!!");        
}

jsFiddle

链接地址: http://www.djcxy.com/p/2378.html

上一篇: 检查元素是否在屏幕上可见

下一篇: 检查元素在DOM中是否可见