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js跳转页面方法实现汇总

2014-02-11 00:00 726 查看
<span id="tiao">3</span><a href="javascript:countDown"></a>布丁足迹;秒后自动跳转……<meta http-equiv=refresh content=3;url='/search/billsearch.jsp'</ul> 
<!--脚本开始--> 
<script language="javascript" type=""> 
function countDown(secs){ 
tiao.innerText=secs; 
if(--secs>0) 
setTimeout("countDown("+secs+")",1000); 
} 
countDown(3); 
</script> 
<!--脚本结束--> 

按钮式: 

<INPUT name="pclog" type="button" value="GO" onClick="location.href='http://www.ddhbb.com/'"> 

链接式: 

<a href="javascript:history.go(-1)">返回上一步</a> 
<a href="<%=Request.ServerVariables("HTTP_REFERER")%>">返回上一步</a> 

直接跳转式: 

<script>window.location.href='http://www.ddhbb.com';</script> 

开新窗口: 

<a href="javascript:" onClick="window.open('http://www.ddhbb.com/blog/guestbook.asp','','height=500,width=611,scrollbars=yes,status=yes')">布丁足迹</a>

JS跳转页面参考代码

第一种:
<script language="javascript" type="text/javascript">
window.location.href="login.jsp?backurl="+window.location.href;
</script>
第二种:
<script language="javascript">
alert("返回");
window.history.back(-1);
</script>
第三种:
<script language="javascript">
window.navigate("top.jsp");
</script>
第四种:
<script language="JavaScript">
self.location='top.htm';
</script>
第五种:
<script language="javascript">
alert("非法访问!");
top.location='xx.jsp';
</script>

=====javascript中弹出选择框跳转到其他页面=====
<script language="javascript">
<!--
function logout()...{
if (confirm("你确定要注销身份吗?是-选择确定,否-选择取消"))...{
window.location.href="logout.asp?act=logout"
}
}
-->
</script>

=====javascript中弹出提示框跳转到其他页面=====
<script language="javascript">
<!--
function logout()...{
alert("你确定要注销身份吗?");
window.location.href="logout.asp?act=logout"
}
-->
</script>
[/code]
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