javascript 如何使弹出窗口始终在顶部
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How to make Popup window always on top
提问by user2854563
I want to open a popup window which should be on top when ever a link from its parentpage is clicked. For example when I click on a link (on a parent page) a popup window open and when I click on another link of parent page the previous popup window become on top of that parent page (and all other open browser's windows)
我想打开一个弹出窗口，当parent点击其页面上的链接时，该窗口应该位于顶部。例如，当我点击一个链接（在父页面上）一个弹出窗口打开时，当我点击父页面的另一个链接时，上一个弹出窗口成为该父页面（以及所有其他打开的浏览器窗口）的顶部
I have tried the following codes in that popup window HTML but with no luck.
我在弹出窗口 HTML 中尝试了以下代码，但没有成功。
<body onLoad="window.focus();">
and
和
<body onLoad="self.focus();">
and
和
<body onBlur="self.focus();">
Neither of these works. I am using Chrome browser if that helps. Please help!
这些都不起作用。如果有帮助，我正在使用 Chrome 浏览器。请帮忙！
回答by Imran Omer
Try adding .focus()at the end of your popup window javascript (just before ;)
尝试.focus()在弹出窗口的末尾添加javascript（就在 之前;）
