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我有弹出窗口,当您进入我的网站时显示 - 这是一个警告标签。我只希望它出现在第一次访问,而不是再次。目前每次点击刷新或返回主页时都会显示。显示弹出式菜单一次性
下面是JavaScript代码(只要有人点击了 '警告按钮输入'):
<script>
"use strict"
warning_popup();
function warning_popup() {
addEventListener('load', start);
function start() {
// popup block background
var bkg = document.createElement("div");
bkg.id = "warning-background";
document.body.insertBefore(bkg, document.body.firstChild);
// popup window
var box = document.createElement("div");
box.id = "warning-window";
document.getElementById("warning-background").appendChild(box);
// warning title
var title = document.createElement("div");
title.id = "warning-title";
title.className = "page-title-wrapper page-title";
title.innerHTML = "<h1>Binge Eating Disorder<\h1>";
document.getElementById("warning-window").appendChild(title);
// warning description
var desc = document.createElement("div");
desc.id = "warning-desc";
desc.className = "page-desc";
desc.innerHTML = "<p>Binge Eating Disorder is disease that I take very seriously.<p>";
document.getElementById("warning-window").appendChild(desc);
// warning button enter
var enter = document.createElement("div");
enter.id = "warning-enter";
enter.className = "page-desc";
enter.innerHTML = "<p>View</p>";
document.getElementById("warning-window").appendChild(enter);
// warning button back
//var back = document.createElement("div");
//back.id = "warning-back";
// back.className = "page-desc";
// back.innerHTML = "<p>Take Me Back</p>";
// document.getElementById("warning-window").appendChild(back);
// listens for button clicks
document.querySelector("#warning-enter").addEventListener("click", function() {
document.querySelector("#warning-background").style.visibility = "hidden";
});
document.querySelector("#warning-back").addEventListener("click", function() {
window.history.back();
});
}
}
解决方法很简单只是设置有失效之日起一年一个javascript的cookie从现在每次检查页面加载。如果该cookie存在,则不运行show_popup函数 –