单击书签链接后打开弹出窗口 [英] open pop up after clicking bookmark link

问题描述

我正在处理一个简短的申请.在其中有后续内容.

I am working on a short application. In which I have followiing things.

1. 将链接从我的页面移动到浏览器的书签工具栏-完成.
2. 单击书签链接后，应打开弹出窗口.(用户可以在任何其他网站上).他将点击书签链接，然后弹出来打开-未完成
3. 该弹出窗口应获取父窗口的标题，元数据，描述(用户当前打开的网站).--完成

很明显，我的问题是关于 2分.我不知道如何使此书签链接到绝对的CSS/JavaScript，以便它可以在任何网站上打开.

Obviously My question is about 2 Point. I cant figure out how to make this bookmark link to an absolute css/javascript so that it can open on any website.

推荐答案

您可以添加将JavaScript链接到网址，如下所示:

You can add javascript to URLs like so:

<a href="javascript:alert('hello');">Bookmarklet</a>

请注意，我在URL前面加上了 javascript:，以指示浏览器应以javascript格式执行URL.

Note that I have prepended the URL with javascript: to indicate the browser that it should execute the URL as javascript.

所以对于流行音乐，它就像:

So for a pop it would be something like:

<a href="javascript:window.open('http://www.example.org');">Bookmarklet</a>

此处是一个jsFiddle ，上面的代码供您测试.

Here is a jsFiddle with the code above for you to test.

有关小书签的更多信息，您可能喜欢以下文章: http://betterexplained.com/articles/how-to-make-a-bookmarklet-for-your-web-application/

For more information about bookmarklets you might like the following article: http://betterexplained.com/articles/how-to-make-a-bookmarklet-for-your-web-application/

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