HTML5历史记录：如何查看弹出的URL（即单击“后退”按钮时所处的页面）
[英]HTML5 history: how can I see the URL that was popped (ie, the page we were on when we hit the back button)
问题描述
I am using the HTML5 history API 
我正在使用HTML5历史记录API
I notice that if: 我注意到，如果：
- if I am on a page called /first 如果我在名为/ first的页面上
- I use pushState to /second 我使用pushState到/ second
- And then hit the back button 然后点击后退按钮
The event handler for window.onpopstate() returns the state for /first, as documented. 如前所述，window.onpopstate（）的事件处理程序返回/ first的状态。
In order to transition from /second to /first, I would like to see the URL that was popped - ie, /second in this case . 为了从/ second过渡到/ first， 我想查看弹出的URL，即本例中的/ second 。 What's the best way of doing that? 最好的方法是什么？
1 个解决方案
解决方案1
1 已采纳 2012-04-16 14:43:40
In lieu of a proper answer, I've implemented the following workaround: 代替正确的答案，我实现了以下解决方法：
- Using a
data-urlhackattribute on the<body>element when a page is loaded加载页面时在<body>元素上使用data-urlhack属性 - In
onpopstate()handler, checking the value of this attribute so I can transition from this page (where the back button was hit) to the page the back button has sent us to.在onpopstate()处理程序中，检查此属性的值，以便我可以从此页面（单击后退按钮的位置）过渡到后退按钮将我们发送到的页面。
Also ensure that pages loaded in onpopstate() don't try and push use pushState() themselves. 还要确保加载到onpopstate()中的页面不要尝试自己推送使用pushState() 。
声明:本站的技术帖子网页，遵循CC BY-SA 4.0协议，如果您需要转载，请注明本站网址或者原文地址。任何问题请咨询:yoyou2525@163.com.