Question

此脚本会旋转5个弹出式标签，每次网页手动刷新时都会显示一个弹出式按钮。我希望脚本每60秒自动旋转一个弹出式标签。

如果某个天才能成功，我会非常感激他

此致

<script language="JavaScript">
<!--  
var frequencyCap = 12; 
function setCookie(cookieName,cookieValue, expirehours) {
  if (frequencyCap > 0) {
    var today = new Date();
    var expire = new Date();
    expire.setTime(today.getTime() + 10000 * frequencyCap);
    document.cookie = cookieName+"="+escape(cookieValue) + ";expires="+expire.toGMTString();
  } else {
    document.cookie = cookieName+"="+escape(cookieValue);
  }
}
function ReadCookie(cookieName) {
 var theCookie=""+document.cookie;
 var ind=theCookie.indexOf(cookieName);
 if (ind==-1 || cookieName=="") return ""; 
 var ind1=theCookie.indexOf(';',ind);
 if (ind1==-1) ind1=theCookie.length; 
 return unescape(theCookie.substring(ind+cookieName.length+1,ind1));
}
if (ReadCookie('cookie1') != '1') { 
 setCookie('cookie1','1', frequencyCap); 
document.write("TAG POPUP-1");
}else if (ReadCookie('cookie2') != '1') { 
 setCookie('cookie2','1', frequencyCap); 
document.write("TAG POPUP-2");
}else if (ReadCookie('cookie3') != '1') { 
 setCookie('cookie3','1', frequencyCap); 
document.write("TAG POPUP-3");
}else if (ReadCookie('cookie4') != '1') { 
 setCookie('cookie4','1', frequencyCap); 
document.write("TAG POPUP-4");
}else if (ReadCookie('cookie5') != '1') { 
 setCookie('cookie5','1', frequencyCap); 
document.write("TAG POPUP-5");
}
// --> 
</script>

Answer 1

这是一个简单的计时器功能。您需要做的只是将代码放入切换弹出标签

var timer = setInterval(rotate, 60000);
var idx = 0;

function rotate() {
    if (idx++ < 5) {
    //Insert code to switch popup tags
    document.getElementById("demo").innerHTML = "Index: " + idx;
    //  ^ demo of timer working
    if (idx >= 5)
        idx = 0;
    }
}

https://jsfiddle.net/3t7nqpg7/1/

为此POPUP javascript添加定时10秒的旋转器功能

1 个答案: