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分享在HTML页面上document对象的onkeydown与onkeyup事件不能同时触发？
<html>
<head>
<meta http-equiv="Content-Language" content="zh-cn">
<meta http-equiv="Content-Type" content="text/html; charset=gb2312">
<title>New Page 1</title>
<script language=javascript for="document" event="onkeyup">
alert("keyUp");
</script>
<script language=javascript for="document" event="onkeydown">
alert("keyDown");
</script>
<script language=javascript for="document" event="onkeypress">
alert("keyPress");
</script>
</head>
<body>
Hello
</body>
</html>
<head>
<meta http-equiv="Content-Language" content="zh-cn">
<meta http-equiv="Content-Type" content="text/html; charset=gb2312">
<title>New Page 1</title>
<script language=javascript for="document" event="onkeyup">
alert("keyUp");
</script>
<script language=javascript for="document" event="onkeydown">
alert("keyDown");
</script>
<script language=javascript for="document" event="onkeypress">
alert("keyPress");
</script>
</head>
<body>
Hello
</body>
</html>
...全文
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saintKnight 2003-07-14
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up
saintKnight 2003-07-10
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我是想在事件onkeydown与onkeyup中分别调用别的方法，但onkeyup事件中的方法根本就没执行
saintKnight 2003-07-10
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为什么使用alert()方法不行？
ydr2002 2003-07-10
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向kingdomzhf() 学习
kingdomzhf 2003-07-10
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<html>
<head>
<meta http-equiv="Content-Language" content="zh-cn">
<meta http-equiv="Content-Type" content="text/html; charset=gb2312">
<title>New Page 1</title>
</head>
<body>
<script FOR="document" EVENT="onkeyup()">
status = 'keyup';
</script>
<script FOR="document" EVENT="onkeydown()">
status='keydown';
</script>
<script FOR="document" EVENT="onkeypress()">
status='keypress';
</script>
Hello
</body>
</html>
keyup触发了
你看状态栏
不要用alert()
<head>
<meta http-equiv="Content-Language" content="zh-cn">
<meta http-equiv="Content-Type" content="text/html; charset=gb2312">
<title>New Page 1</title>
</head>
<body>
<script FOR="document" EVENT="onkeyup()">
status = 'keyup';
</script>
<script FOR="document" EVENT="onkeydown()">
status='keydown';
</script>
<script FOR="document" EVENT="onkeypress()">
status='keypress';
</script>
Hello
</body>
</html>
keyup触发了
你看状态栏
不要用alert()
