有没有办法让我的页面使用Javascript实时监控cookie,如果设置了cookie,那么它会发送警报吗?
我不明白的部分是如何实时检查cookie?例如,用户可能会单击一个复选框,将设置一个cookie,然后将页面设置为"请参阅"现在已经设置了cookie并触发警报,所有这些都是实时的。
以下是一个示例(未正确编写),它显示了我的目标。
<div id="click" onclick="setcookie()"></div>
<script>
function checkcookie() {
if (cookie_exists)
alert('cookie exists')
}
</script>
我不明白的部分是如何让checkcookie()函数持续监控我刚刚点击设置的cookie的存在?
答案 0 :(得分:0)
以下是编写和阅读Cookie的基本功能:
function setCookie(cname, cvalue, exdays) {
var d = new Date();
d.setTime(d.getTime() + (exdays * 24 * 60 * 60 * 1000));
var expires = 'expires=' + d.toUTCString();
document.cookie = cname + '=' + cvalue + ';' + expires + '';
path = '/';
}
function getCookie(cname) {
var name = cname + '=';
var ca = document.cookie.split(';');
for (var i = 0; i < ca.length; i++) {
var c = ca[i];
while (c.charAt(0) == ' ') {
c = c.substring(1);
}
if (c.indexOf(name) == 0) {
return c.substring(name.length, c.length);
}
}
return null;
}
function checkCookie(cname) {
var cookie = getCookie(cname);
if (cookie != null) {
return true;
} else {
return false;
}
}
function remove_cookie(cname) {
document.cookie = cname + '=;expires=Thu, 01 Jan 1970 00:00:00 UTC;path=/;'
}
来自w3schools.com/js/js_cookies.asp
答案 1 :(得分:0)
唯一的方法是使用
window.setInterval("CheckCookie()", 1000);
每秒检查一次cookie
答案 2 :(得分:0)
您必须实施间隔超时以持续检查cookie的值。
function getCookie(name) {
var dc = document.cookie;
var prefix = name + "=";
var begin = dc.indexOf("; " + prefix);
if (begin == -1) {
begin = dc.indexOf(prefix);
if (begin != 0) return null;
}
else
{
begin += 2;
var end = document.cookie.indexOf(";", begin);
if (end == -1) {
end = dc.length;
}
}
// because unescape has been deprecated, replaced with decodeURI
//return unescape(dc.substring(begin + prefix.length, end));
return decodeURI(dc.substring(begin + prefix.length, end));
}
// Method used for continuously monitoring a cookie
function monitorCookie(cookieName) {
setInterval(function() {
var cookieValue = getCookie(cookieName);
console.log('Yummy Cookie =' + cookieValue);
}, 500);// monitor cookie every 500 miliseconds.
}
$(document).ready(function() {
monitorCookie("YummyCookie");
});