问题描述
我刚开始反应并在组件中包含此链接:
I'm new to react and have this link in a component:
<a href="https://example.com/faq.html"> FAQ </a>
我想在响应之外处理 faq.html.
I want to server faq.html outside react.
问题是反应将链接视为内部链接并给出404.
The problem is that react treats the link as internal and gives 404.
我看到了一个类似的问题，建议使用 <Route ...，但后来我不知道如何将超链接转换为路由.
I have seen a similar question which suggest to use <Route ..., but then I don't know how to convert the hyperlink to a Route.
我也知道我可以将 target="_blank" 添加到标签中，但这不是我理想的解决方案.
Also I know that I can add target="_blank" to the tag, but that's not my ideal solution.
非常感谢您帮助解决这个问题.
So appreciate your help to solve this.
推荐答案
你想多了.您可以像使用 HTML 一样执行此操作.这就是 JSX 的重点.你不需要路由.在此上下文中，外部链接不是路由.也许您弄错了一些周围的代码，但从您的示例中看不到这一点.
You're overthinking this. You can do it just like you would with HTML. That's the point of JSX. You don't need Route. An external link isn't a route in this context. Maybe you got some surrounding code wrong but can't see that from your example.
如果您想包含其他组件，您需要将它们包装在所有一个组件中.例如，使用Div"或片段<></>.否则，您可以像使用 HTML 一样执行此操作.
If you want to include other components you need to wrap them in all one component. For example with 'Div' or a fragment <></>. Otherwise, you can do it just as you would with HTML.
*rel="noreferrer";出于安全考虑，建议使用.
*rel="noreferrer" recommended for security.
<a href="https://example.com/faq.html" rel="noreferrer">
FAQ
</a>
这篇关于如何在反应中制作外部超链接?的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持！