我正在创建一个JavaScript弹出窗口。代码如下。
HTML:
<div id="ac-wrapper" style='display:none' onClick="hideNow(event)">
<div id="popup">
<center>
<h2>Popup Content Here</h2>
<input type="submit" name="submit" value="Submit" onClick="PopUp('hide')" />
</center>
</div>
</div>
CSS:
#ac-wrapper {
position: fixed;
top: 0;
left: 0;
width: 100%;
height: 100%;
background: url("images/pop-bg.png") repeat top left transparent;
z-index: 1001;
}
#popup {
background: none repeat scroll 0 0 #FFFFFF;
border-radius: 18px;
-moz-border-radius: 18px;
-webkit-border-radius: 18px;
height: 361px;
margin: 5% auto;
position: relative;
width: 597px;
}
剧本:
function PopUp(hideOrshow) {
if (hideOrshow == 'hide') document.getElementById('ac-wrapper').style.display = "none";
else document.getElementById('ac-wrapper').removeAttribute('style');
}
window.onload = function () {
setTimeout(function () {
PopUp('show');
}, 0);
}
function hideNow(e) {
if (e.target.id == 'ac-wrapper') document.getElementById('ac-wrapper').style.display = 'none';
}
jsFiddle链接:
http://jsfiddle.net/K9qL4/2/
问题:
上面的脚本可以正常工作,但是我需要使popUp在我的页面上仅出现一次。
即,当用户关闭弹出窗口时,直到用户重新启动浏览器或清除其缓存/ Cookie后,弹出窗口才应显示。
我尝试使用下面的cookie脚本,但是它对我不起作用。
<SCRIPT LANGUAGE="JavaScript">
<!-- Begin
var expDays = 1; // number of days the cookie should last
var page = "myPage.html";
var windowprops = "width=300,height=200,location=no,toolbar=no,menubar=no,scrollbars=no,resizable=yes";
function GetCookie (name) {
var arg = name + "=";
var alen = arg.length;
var clen = document.cookie.length;
var i = 0;
while (i < clen) {
var j = i + alen;
if (document.cookie.substring(i, j) == arg)
return getCookieVal (j);
i = document.cookie.indexOf(" ", i) + 1;
if (i == 0) break;
}
return null;
}
function SetCookie (name, value) {
var argv = SetCookie.arguments;
var argc = SetCookie.arguments.length;
var expires = (argc > 2) ? argv[2] : null;
var path = (argc > 3) ? argv[3] : null;
var domain = (argc > 4) ? argv[4] : null;
var secure = (argc > 5) ? argv[5] : false;
document.cookie = name + "=" + escape (value) +
((expires == null) ? "" : ("; expires=" + expires.toGMTString())) +
((path == null) ? "" : ("; path=" + path)) +
((domain == null) ? "" : ("; domain=" + domain)) +
((secure == true) ? "; secure" : "");
}
function DeleteCookie (name) {
var exp = new Date();
exp.setTime (exp.getTime() - 1);
var cval = GetCookie (name);
document.cookie = name + "=" + cval + "; expires=" + exp.toGMTString();
}
var exp = new Date();
exp.setTime(exp.getTime() + (expDays*24*60*60*1000));
function amt(){
var count = GetCookie('count')
if(count == null) {
SetCookie('count','1')
return 1
}
else {
var newcount = parseInt(count) + 1;
DeleteCookie('count')
SetCookie('count',newcount,exp)
return count
}
}
function getCookieVal(offset) {
var endstr = document.cookie.indexOf (";", offset);
if (endstr == -1)
endstr = document.cookie.length;
return unescape(document.cookie.substring(offset, endstr));
}
function checkCount() {
var count = GetCookie('count');
if (count == null) {
count=1;
SetCookie('count', count, exp);
window.open(page, "", windowprops);
}
else {
count++;
SetCookie('count', count, exp);
}
}
// End -->
</script>
最佳答案
我在这种情况下最好使用localStorage而不是cookie。
localStorage具有更直观的界面,用户无法限制
要使用此功能。我已经更改了您的代码。
function PopUp(hideOrshow) {
if (hideOrshow == 'hide') {
document.getElementById('ac-wrapper').style.display = "none";
}
else if(localStorage.getItem("popupWasShown") == null) {
localStorage.setItem("popupWasShown",1);
document.getElementById('ac-wrapper').removeAttribute('style');
}
}
window.onload = function () {
setTimeout(function () {
PopUp('show');
}, 0);
}
function hideNow(e) {
if (e.target.id == 'ac-wrapper') document.getElementById('ac-wrapper').style.display = 'none';
}
这是工作的jsFiddle。
http://jsfiddle.net/zono/vHG7j/
最好的祝福。
关于javascript - 使我的javaScript弹出窗口仅出现一次,我们在Stack Overflow上找到一个类似的问题:https://stackoverflow.com/questions/23775440/