有没有一种简单的方法可以在JSF中实现路由?

问题描述：

我正在尝试实现一个通用"视图，其中(部分)显示的内容取决于URL.例如

I am trying to implement a "generic" view where (part of) the content displayed depends on the URL. E.g.

如果/somepath/somepage.xhtml指向不存在的文件，而不是直接出现404错误，我想尝试使用通用视图/genericview.xhtml从数据库中检索/somepath/somepage.xhtml的内容，在该视图中我有一些东西像:

If /somepath/somepage.xhtml points to a non existing file, instead of going straight to a 404 error I want to try to retrieve /somepath/somepage.xhtml's content from the database using a generic view, /genericview.xhtml, where I have something like:

<h:outputText value="#{genericViewBean.content_lg}"
                escape="false" />

(如果由后备bean找到)，它将根据最初请求的viewId从tgenericcontent表中输出数据库条目的内容:

which, if found by the backing bean, would output the content of the database entry from a tgenericcontent table, depending on the originally requested viewId:

 webpath                              | content
 /somepath/somepage.xhtml             | <p>This is a test</p>
 /someotherpath/someotherpage.xhtml   | <p>A different test</p>

如果在该表中未找到视图内容，则将返回标准404错误.

If the view content is not found in that table then the standard 404 error would be returned.

我最接近的浪费是克隆/genericview.xhtml，仅更改文件路径(例如，更改为/somepath/somepage.xhtml).但这给我每个视图一个文件的精确副本，这非常混乱，而且它不允许我仅通过向数据库添加一个条目来创建新的URL.

The closest I got wast to clone /genericview.xhtml changing only the file path (for example, to /somepath/somepage.xhtml). But that gets me one exact copy of the file per view, it is quite messy, and it doesn't allow me to create a new url just by adding an entry to my database.

如何在不克隆/genericview.xhtml的情况下获得相同的结果?

How can I get the same result without cloning /genericview.xhtml?

(PS:我已经读过有关prettyfaces的内容，但是没有更简单的解决方案吗?)

(P.S: I have read about prettyfaces, but isn't there a simpler solution?)

答

为此，通常 servlet过滤器正在被使用. PrettyFaces ， FacesViews 也是这样做的.

For that, normally a servlet filter is being used. PrettyFaces, UrlRewriteFilter and FacesViews also do it that way.

您可以通过 ServletContext#getResource() ，它将在不存在的资源上返回null.如果资源存在，则只需通过 RequestDispatcher#forward() .

You can get the request URI by HttpServletRequest#getRequestURI(). You can check the existence of a web resource by ServletContext#getResource() which will return null on non-existent resources. If the resource exists, just continue the request by FilterChain#doFilter(), else forward the request to the generic view by RequestDispatcher#forward().

总而言之，这就是过滤器的外观:

All in all, this is how the filter could look like:

@WebFilter("/*")
public class GenericViewFilter implements Filter {

    @Override
    public void doFilter(ServletRequest req, ServletResponse res, FilterChain chain) throws IOException, ServletException {
        HttpServletRequest request = (HttpServletRequest) req;
        HttpServletResponse response = (HttpServletResponse) res;
        String relativeRequestURI = request.getRequestURI().substring(request.getContextPath().length());

        boolean resourceExists = request.getServletContext().getResource(relativeRequestURI) != null;
        boolean facesResourceRequest = request.getRequestURI().startsWith(request.getContextPath() + ResourceHandler.RESOURCE_IDENTIFIER));

        if (resourceExists || facesResourceRequest) {
            chain.doFilter(request, response);
        }
        else {
            request.getRequestDispatcher("/genericview.xhtml").forward(request, response);
        }
    }

    // ...
}

在/genericview.xhtml中，原始请求URI可用作为以

In the /genericview.xhtml, the original request URI is available as request attribute keyed with RequestDispatcher#FORWARD_REQUEST_URI. You could use it in @PostConstruct of backing bean associated with the view in order to pull the right content from the DB.

String originalRequestURI = (String) externalContext.getRequestMap().get(RequestDispatcher.FORWARD_REQUEST_URI);
// ...

有没有一种简单的方法可以在JSF中实现路由?

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