How to get .outerHeight of element if visible

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Im woundering if its possible to get .outerHeight() of element only if its is visible

I currently have

var $viewItem = $('.test').find('.item');


var viewItemHeight = $viewItem.outerHeight(true);

The problem is $viewItem

Will return two elements at a time, one will be hidden. If is possible to test and only get the height of the visible one ?

  1. Post:183

    Re: How to get .outerHeight of element if visible

    What about the visible selector?

    var $viewItem = $('.test').find('.item:visible');

    var viewItemHeight = $viewItem.outerHeight(true);


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