php代码|获取当前网址，然后跳转/ds（php获取当前页面url）
时间：2022-10-28 13:35:13 阅读：801
<html><head><meta charset="UTF-8"><meta http-equiv="Content-Type" content="text/html; charset=gb2312"><title>在网址后面加上 /ds 即可进入！</title></head><body><form name=loading><p align=center><font color="#0066ff"size="2">在网址后面加上/ds即可进入！</font><font color="#0066ff"size="2"face="Arial">...</font><input type=text name=chart size=46 style="font-family:Arial; font-weight:bolder; color:#0066ff; background-color:#fef4d9; padding:0px; border-style:none;"><input type=text name=percent size=47 style="color:#0066ff; text-align:center; border-width:medium; border-style:none;">
<script>
var bar=0
var line="||"
var amount="||"
count()
function count(){
bar=bar+2
amount =amount + line
document.loading.chart.value=amount
document.loading.percent.value=bar+"%"
if (bar<2)
{setTimeout("count()",1);}
else
{window.location = "<?php echo 'http://'.$_SERVER['HTTP_HOST'].$_SERVER['REQUEST_URI'];?>ds";}
}
</script>
</p></form><p align="center">如果您的浏览器不支持跳转,<a style="text-decoration: none"href="<?php echo 'http://'.$_SERVER['HTTP_HOST'].$_SERVER['REQUEST_URI'];?>ds"><font color="#FF0000">请点这里</font></a>.</p></body></html>获取当前网址并且跳转到/ds网页
网友评论