Check visibility of an object with JavaScript_问答_开发者

Check visibility of an object with JavaScript

开发者 https://www.devze.com 2023-02-06 10:55 出处：网络

I have a variable called \"object\". How c开发者_开发百科an I check with JavaScript if it is visible?

相关专题：css javascript

I have a variable called "object". How c开发者_开发百科an I check with JavaScript if it is visible?

I tried !object.getAttribute("dislay", "none")... but that doesn't work.

Could somebody help me please?

Thank you!

function isvisible(obj) {
  return obj.offsetWidth > 0 && obj.offsetHeight > 0;
}

as it's implemented in JQuery.

If you use jQuery, the following will return true if the object is visible:

$(object).is(':visible');

If not, you can try these:

if (object.style['display'] != 'none')

But this will work only if display:none has been set explicitly on this object.

To get value of a display style using javascript you can use:

IE:

document.getElementById('mydiv').currentStyle.display
or
object.currentStyle.display

Others:

document.getElementById('mydiv').getComputedStyle('display')
or
object.getComputedStyle('display')

if (object.style.visibility <> 'visible' || object.style.display == 'none')

If it doesn't work, try to use

 if (object.currentStyle.visibility <> 'visible' || object.currentStyle.display == 'none')

Doesn't look like you're using the getAttribute method correctly. Try taking a look at this.

Here is the working version: http://jsfiddle.net/PEA4j/

<html>
<head>
    <script src="http://ajax.aspnetcdn.com/ajax/jquery/jquery-1.4.4.min.js" type="text/javascript"></script>
    <script type="text/javascript">
        $(function () {
            alert("Is #foo1 visible: " + $("#foo1").is(":visible") + "\nIs #foo2 visible: " + $("#foo2").is(":visible"));

        });
    </script>
</head>
<body>
<div id="foo1" style="display:none">foo1 display none</div>
<div id="foo2">foo2 no display property;</div>
</body>
</html>