我刚刚在使用Javascript和我在页面加载事件的麻烦。提交表格1后,应出现表格2。Javascript body onload
<html>
<head>
<script src="jquery1.9.1.js" type="text/javascript"></script>
<link rel="stylesheet" type="text/css" href="sCss.css">
</head>
<script type="text/javascript">
function hide(obj1) {
obj1 = document.getElementById(obj 1);
if (document.cookie.indexOf('mycookie') == -1) {
document.cookie = 'mycookie = 1';
obj1.style.display = 'none';
} else {
obj1.style.display = 'block';
}
}
function show(obj2) {
var cookie = "test"
obj2 = document.getElementById(obj2);
obj2.style.display = 'block';
document.cookie = 'mycookie = 1';
}
</script>
<body onload="hide('form2')">
<form id="form1" onsubmit="show('form2')">
<table id="table1" border="1" >
<tr>
<td>
<input type="text" name="txt1" id="txt1" />
</td>
<td>
<input type="submit" name="submit1" id="submit1" />
</td>
</tr>
</table>
</form>
<form id="form2">
<table id="table2" border="1" >
<tr>
<td>
<input type="text" name="txt2" id="txt2" />
</td>
<td>
<input type="submit" name="submit2" id="submit2" />
</td>
</tr>
</table>
</form>
</body>
</html>
根据给出的代码。点击form1的提交按钮后,form2出现并立即消失。
你真正想要提交表单?或者您是否想要显示下一部分信息要完成? –
@johnGerdsen tnx询问..在我的情况下,他们需要提交表单,然后另一种形式将弹出,btw我张贴的代码只是一个原型,因为我无法妥善处理onload事件:( – zxc
'提交'意味着一些特殊的你只是想隐藏form1并显示form2? – sync