Javascript 打开弹出窗口，单击链接，在父窗口中打开，关闭弹出窗口？
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Open popup, click link, open in parent window, close popup?
提问by Adam
I need help please. What I want to do from my page is to open a popup for which I use this:
我需要帮助。我想从我的页面做的是打开一个弹出窗口，我使用它：
<a class="txt-button" onclick="javascript:void window.open('http://mypage.com/1/index.html','','width=700,height=500,resizable=false,left=0,top=0');return false;">Buy Now</a>
When a link on the popup is clicked I want it to open in the main window and for the popup to close.
单击弹出窗口上的链接时，我希望它在主窗口中打开并关闭弹出窗口。
I've tried many things and cant get it to work:-(
我已经尝试了很多东西，但无法让它工作:-(
回答by Arun P Johny
回答by TildalWave
To access main openerwindow from a pop-upuse the window.openerobject. You can access any opener property, or even a function call like that (as long as domains match, as @N Rohler pointed out in the comment to your question), for example to navigate use window.opener.location.href. Then you close your pop-up with window.close();like this:
要从弹出opener窗口访问主窗口，请使用该window.opener对象。您可以访问任何 opener 属性，甚至是这样的函数调用（只要域匹配，正如@N Rohler 在对您的问题的评论中指出的那样），例如导航 use window.opener.location.href。然后你window.close();像这样关闭你的弹出窗口：
<a href="JavaScript:void(0);" onclick="openInParent('http://example.com/');">
click me
</a>
<script>
function openInParent(url) {
window.opener.location.href = url;
window.close();
}
</script>
