如何按顺序关闭最后打开的弹出窗口?
下面的图片是一个图像，每当弹出窗口打开时，弹出标签就堆叠在一起。
输入图片描述
下面是应用的源代码。
/* ESC popup close start */
$(document).on('keyup', function(e) {
if (e.key == "Escape") $('.window .close').last().click();
});
/* ESC popup close end */
- 确保
close点击事件先于点击触发 - 你需要
addClass到最后一个.window元素，以避免它，当你再次订购最后一个项目
$(".close").on("click" , function(){
$(this).closest(".window").hide();
});
/* ESC popup close start */
$(document).on('keyup', function(e) {
if (e.key == "Escape") $('.window:not(.hidden)').last().addClass("hidden").find(".close").click();
});
/* ESC popup close end */<script src="https://cdnjs.cloudflare.com/ajax/libs/jquery/3.3.1/jquery.min.js"></script>
<div class="window">Window 1<div class="close">X</div></div>
<div class="window">Window 2<div class="close">X</div></div>
<div class="window">Window 3<div class="close">X</div></div>只需将弹出窗口设置为任意变量，然后使用window.close关闭该变量
const daddyWindow = window.open("", "window" + number, "width=200,height=100");
daddyWindow.document.write('window Popup code');
daddyWindow.close();
试试这个:
<html>
<header>
<script src="https://cdnjs.cloudflare.com/ajax/libs/jquery/3.3.1/jquery.min.js"></script>
<script>
let openedWindows = [];
let newWindow;
let number = 0;
function openWindow() {
number++;
newWindow = window.open("", "window" + number, "width=200,height=100");
newWindow.document.write(`<a href="javascript:window.close();"class="video-close">Close this window ${number}</a>`);
openedWindows.push(newWindow);
}
$(document).on('keyup', function (e) {
if (e.key == "Escape") {
if (openedWindows && openedWindows.length >= 0 && openedWindows[openedWindows.length - 1]) {
openedWindows[openedWindows.length - 1].close();
openedWindows.pop()
}
}
});
</script>
</header>
<body>
<h3 style="color:brown"> Close Window Example </h3>
<button onclick="openWindow()">Open New Window</button>
</body>
</html>