Javascript：如何引用通过发布打开的子窗口

发布于 2024-12-12 04:13:27 字数 375 浏览 0 评论 0原文

我想引用我打开的子窗口，以便我可以从父窗口调用它的一些方法。为此，我需要保留对窗口的引用。我找到的所有示例都保留引用：

var windowRef = window.open(...);

但我不使用 window.open。我提交表单是因为我需要向其中发布数据：

var frmMain = document.forms['frmMain'];
frmMain.action = "http://www.somedomain.com";
link.attr("href");
frmMain.target = "_blank";
frmMain.submit();

如何使用帖子引用打开的窗口？谢谢！

原文

I want to reference the child window that I have opened so that I can call some method on it from the parent. To do that I need to keep a reference to the window. All the examples I have found keep the reference by:

var windowRef = window.open(...);

But I don't use window.open. I submit the form because I need to post data to it:

var frmMain = document.forms['frmMain'];
frmMain.action = "http://www.somedomain.com";
link.attr("href");
frmMain.target = "_blank";
frmMain.submit();

How can I reference the open window using a post? Thanks!

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若有似无的小暗淡 2024-12-19 04:13:27

只要您的 target 属性设置为真实名称，窗口对象的 open 方法实际上可以为您创建引用。

var frmMain = document.forms['frmMain'];
frmMain.action = "http://www.somedomain.com";
link.attr("href");
frmMain.target = "myawsmwindow";
frmMain.submit();

var myawsmwindow = window.open("", "myawsmwindow"); // gets you a variable ref to the named window.

编辑：这是一个演示 jsfiddle。

The open method of the window object can actually create the reference for you as long as your target attribute is set to a real name.

var frmMain = document.forms['frmMain'];
frmMain.action = "http://www.somedomain.com";
link.attr("href");
frmMain.target = "myawsmwindow";
frmMain.submit();

var myawsmwindow = window.open("", "myawsmwindow"); // gets you a variable ref to the named window.

edit: Here is a demonstration jsfiddle.

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