[英]Why it doesn't work, style.display
I have a short script. 我有一个简短的剧本。 I want to check the visibility oder the Display of a div.
我想检查div显示的可见性。 But it doesn't work.
但这是行不通的。 And i have no idea why, i found some other scripts with the same Code.
而且我不知道为什么,我发现了其他一些具有相同代码的脚本。
I didn't get the message JA 我没有收到消息JA
<html>
<head>
<script>
function test()
{
alert('J');
if (document.getElementById('t').style.visibility == "visible")
{
alert('JA');
}
if (document.getElementById('t').style.display == "block")
{
alert('JA');
}
}
</script>
</head>
<body>
<div id = 't'>
test
</div>
<input type = 'button' onClick = 'test();'>
</body>
</html>
The style
property only represents the CSS directives specified in the element's style
attribute therefore, both visibility
and display
will be empty. style
属性仅表示在元素的style
属性中指定的CSS指令,因此visibility
和display
均为空。
What you want is computed style 您想要的是计算样式
var t = document.getElementById('t'),
style = window.getComputedStyle(t);
console.log('visibility', style.visibility);
console.log('display', style.display);
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