According to this question page and the accepted answer here, the right way to get inherited CSS values via Javascript is getComputedStyle(). However, that does not work in the following example:
<!DOCTYPE html>
<html lang="en">
<body>
<form id="iterateThroughMe">
<div class="notHidden"><input name="myNum" type="number" /></div>
<div><input name="myOtherNum" type="number" /></div>
<input name="myText" type="text" />
<div id="hider" style="display: none;">
<input name="hiddenElement" type="number" />
</div>
</form>
<script src="https://ajax.googleapis.com/ajax/libs/jquery/1.12.4/jquery.min.js"></script>
<script>
var inputs = $("#iterateThroughMe").find('input');
for(var i = 0; i<inputs.length; i++) {
var displayStyle = window.getComputedStyle(inputs[i], null).display;
console.log(
(($(inputs[i]).is(":hidden"))?"in":"")+
"visible element: ",inputs[i],"has display style",displayStyle);
}
</script>
</body>
</html>
Saved and opened in a browser, the following is visible on the console:
visible element: <input name="myNum" type="number"> has display style inline-block
visible element: <input name="myOtherNum" type="number"> has display style inline-block
visible element: <input name="myText" type="text"> has display style inline-block
invisible element: <input name="hiddenElement" type="number"> has display style inline-block
There is a specific way to access hidden/visible status in jQuery, noted here, which works in the example above but is not always appropriate. Why does getComputedStyle not distinguish between hidden and non-hidden elements, producing inline-block on all four lines instead of none at the end of the output?
