自定义错误页面-获取最初请求的网址 [英] Custom error page - get originally requested URL

问题描述

问题

这是对昨天(未回答)问题的后续操作(请参阅此处 )，因为我试图找到另一种方法.

This is a follow-up to yesterday's (unanswered) question (see here) as I try to find an alternative approach.

我添加了基本的

    <error-page>  
            <error-code>404</error-code>  
            <location>/404search.jsf</location>  
    </error-page>

..到我的web.xml.现在，我需要获取用户输入的URL才能提交给我的搜索功能，但是我只能设法获取当前URL(在这种情况下为... 404search.jsf)，而不是用户输入的实际查询.

..to my web.xml. I now need to get the URL the user entered to submit to my search function, but I only manage to get the current URL (in this case, ...404search.jsf) instead of the actual query the user entered.

尝试

• HttpServletRequest.getRequestURL返回http://www.website.com/foldername/404search.jsf
• HttpServletRequest.getRequestURI返回/foldername/404search.jsf
• HttpServletRequest.getQueryString完全不返回任何内容

• HttpServletRequest.getRequestURL returns http://www.website.com/foldername/404search.jsf
• HttpServletRequest.getRequestURI returns /foldername/404search.jsf
• HttpServletRequest.getQueryString returns nothing at all

我希望它返回/foldername/wrong-url-the-user-entered#anchor-if-there-is-any

详细信息...

想法是获取用户输入的URL(例如www.site.com/product/99999-product-that-cant-be-found或www.site.com/faq/support-question-that-doesnt-exist)，对其进行正则表达式以删除连字符，并使用99999 product that cant be found或support question that doesnt exist运行搜索查询.

The idea is to get the URL the user entered (such as www.site.com/product/99999-product-that-cant-be-found or www.site.com/faq/support-question-that-doesnt-exist), REGEX it to remove the hyphens and run a search query with 99999 product that cant be found or support question that doesnt exist.

有什么建议吗?

推荐答案

<error-page>位于

The <error-page> is under the covers served by a RequestDispatcher#forward() call. All details of the original request are available as request attribues which are keyed by the keys as identified by RequestDispatcher#FORWARD_XXX constants:

• FORWARD_CONTEXT_PATH :
• FORWARD_PATH_INFO :
• FORWARD_QUERY_STRING :
• FORWARD_REQUEST_URI :
• FORWARD_SERVLET_PATH :

• FORWARD_CONTEXT_PATH: "javax.servlet.forward.context_path"
• FORWARD_PATH_INFO: "javax.servlet.forward.path_info"
• FORWARD_QUERY_STRING: "javax.servlet.forward.query_string"
• FORWARD_REQUEST_URI: "javax.servlet.forward.request_uri"
• FORWARD_SERVLET_PATH: "javax.servlet.forward.servlet_path"

作为初学者，您应该知道所有请求属性都可以通过隐式EL对象 #{requestScope}.

You as starter should know that all request attributes are in EL available via the implicit EL object #{requestScope}.

总而言之，这应该在视图中完成:

So, all with all, this should do in the view:

<p>Unfortunately, the page you requested, #{requestScope['javax.servlet.forward.request_uri']} does not exist</p>

等效地，如果需要的话，这应该在bean中完成:

And equivalently, this should do in the bean, if necessary:

String forwardRequestURI = externalContext.getRequestMap().get(RequestDispatcher.FORWARD_REQUEST_URI);

这篇关于自定义错误页面-获取最初请求的网址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IT屋！