使用javascript将链接标记为已访问 [英] using javascript to mark a link as visited

问题描述

FF2（至少）未标记为链接为：visited如果它触发onclick处理程序而不遵循href。我使用onclick从服务器获取数据，并修改页面和链接样式在这里似乎是适当的。但该链接未标记为已访问。

FF2 (at least) doesn't mark as link as :visited if it triggers the onclick handler without following the href. I'm using onclick to fetch data from a server and modify the page and the link styling seems appropriate here. But the link is not marked as visited.

是否有跨浏览器方式将链接标记为已访问？没有，是否有一种方法来确定浏览器的a：访问的样式并将其应用于链接？

Is there a cross-browser way to mark the link as visited? Failing that, is there a way to determine the browser's a:visited styling and apply it to the link?

感谢所有回复。

Thanks to all who replied.

看起来像答案是：

• 是否有跨浏览器方式将链接标记为已访问？
  
  否，没有办法。如果href在浏览器历史记录中，则链接被标识为已访问。
  
• 有没有办法确定浏览器的a：访问样式？
  
  不，不是通过单独的javascript。
  

• Is there a cross-browser way to mark the link as visited?
  No, there's no way to do this. Links are identified as visited if the href is in the browser history.
• Is there a way to determine the browser's a:visited styling?
  No, not via javascript alone.

推荐答案

这是我怎么做的。仅适用于支持HTML5历史记录API的浏览器。

Here is how I did it. Only works in browsers that support HTML5 history api.

# store the current URL
current_url = window.location.href

# use replaceState to push a new entry into the browser's history
history.replaceState({},"",desired_url)

# use replaceState again to reset the URL
history.replaceState({},"",current_url)

使用replaceState意味着后退按钮不会受到影响。

Using replaceState means that the back button won't be affected.

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