【发布时间】:2015-10-18 06:12:49
【问题描述】:
<!DOCTYPE html>
<html>
<head>
<title>Face</title>
<meta charset="UTF-8">
<script type="text/javascript">
//<![CDATA[
function myFunction(id) {
var e = document.getElementById(id);
if(e.style.visibility == "visible")
e.style.visibility = 'hidden';
else
e.style.visibility = 'visibile';
}
</script>
</head>
<body>
<div style="position: relative; visibility: visible;">
<img src="http://vignette4.wikia.nocookie.net/mrmen/images/5/52/Small.gif/revision/latest?cb=20100731114437"
alt="Pumpkins" id="Pum"/>
<button onclick="myFunction('Pum')">Face</button>
</div>
</body>
</html>
我想要一个按钮来显示/隐藏我的图像。我不明白我做错了什么。我收到一条错误消息,提示“TypeError:无法读取 null 的属性‘可见性’”。如何修复我的错误并让我的程序正常运行?
【问题讨论】:
标签: javascript image show-hide