[英]How to check the existence of Children Window
I have a JSP page A. And on page A, there are some links when click on them, page B or C will be popped up. 我有一个JSP页面A。在页面A上,单击某些链接时,将弹出页面B或C。
Now I have a javascript to run in page A. How can this javascript determine whether there are existing popup page B or C? 现在,我有一个可在页面A中运行的javascript。此javascript如何确定是否存在弹出页面B或C?
First of all you need to ensure that you have references to your opened windows available with you. 首先,您需要确保拥有对打开的窗口的引用。 It is currently not possible to get a list of already open window. 当前无法获得已经打开的窗口的列表。 If you are unsure you can refer to this answer that suggests a decent way of doing this. 如果您不确定,可以参考此答案 , 此举提出了一种不错的方法。
Now, using the window object of the child window, you can check if it is null or not in order to know whether it is open or not: 现在,使用子窗口的窗口对象,可以检查它是否为null以便知道它是否处于打开状态:
var myWindow = window.open("https://www.google.com", "MsgWindow", "width=200,height=100");
function IsWindowClosed (win) {
return win.window == null;
}
console.log("Window is closed: ", IsWindowClosed(myWindow));
Here's a working fiddle: https://jsfiddle.net/6vhgpkmm/1/ 这是一个有效的小提琴: https : //jsfiddle.net/6vhgpkmm/1/
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