如何在Java中使用servlet过滤器来更改传入的servlet请求URL？

Question

如何使用servlet过滤器来更改传入的servlet请求URL

http://nm-java.appspot.com/Check_License/Dir_My_App/Dir_ABC/My_Obj_123

至

http://nm-java.appspot.com/Check_License?Contact_Id=My_Obj_123

？

---

更新:根据BalusC的以下步骤,我想出了以下代码:

public class UrlRewriteFilter implements Filter {

    @Override
    public void init(FilterConfig config) throws ServletException {
        //
    }

    @Override
    public void doFilter(ServletRequest req, ServletResponse res, FilterChain chain) throws ServletException, IOException {
        HttpServletRequest request = (HttpServletRequest) req;
        String requestURI = request.getRequestURI();

        if (requestURI.startsWith("/Check_License/Dir_My_App/")) {
            String toReplace = requestURI.substring(requestURI.indexOf("/Dir_My_App"), requestURI.lastIndexOf("/") + 1);
            String newURI = requestURI.replace(toReplace, "?Contact_Id=");
            req.getRequestDispatcher(newURI).forward(req, res);
        } else {
            chain.doFilter(req, res);
        }
    }

    @Override
    public void destroy() {
        //
    }
}

相关条目web.xml如下所示:

<filter>
    <filter-name>urlRewriteFilter</filter-name>
    <filter-class>com.example.UrlRewriteFilter</filter-class>
</filter>
<filter-mapping>
    <filter-name>urlRewriteFilter</filter-name>
    <url-pattern>/*</url-pattern>
</filter-mapping>

我尝试了服务器端和客户端重定向以及预期的结果.它起作用了,谢谢BalusC!

Answer 1

1. 实施javax.servlet.Filter.
2. 在doFilter()方法中,将传入转换ServletRequest为HttpServletRequest.
3. 用HttpServletRequest#getRequestURI()抢的路径.
4. 使用简单的java.lang.String方法,如substring(),split(),concat()等提取感兴趣的部分,构成了新的路径.
5. 使用任一ServletRequest#getRequestDispatcher(),然后RequestDispatcher#forward()转发该请求/响应新的URL(服务器端重定向,而不是在浏览器地址栏中反射),或投传入ServletResponse到HttpServletResponse,然后HttpServletResponse#sendRedirect()重定向到新的URL的响应(客户端重定向,反映在浏览器地址栏).
6. 注册过滤器web.xml上url-pattern的/*或/Check_License/*根据上下文路径上,或者如果你是在Servlet的3.0已经使用了@WebFilter注解为代替.

如果需要更改URL,请不要忘记在代码中添加一个检查,如果没有,则只需调用FilterChain#doFilter(),否则它将在无限循环中调用自身.

或者,您也可以使用现有的第三方API为您完成所有工作,例如Tuckey的UrlRewriteFilter,它可以像配置Apache一样进行配置mod_rewrite.

Answer 2

您可以使用准备使用Url Rewrite Filter的规则如下:

<rule>
  <from>^/Check_License/Dir_My_App/Dir_ABC/My_Obj_([0-9]+)$</from>
  <to>/Check_License?Contact_Id=My_Obj_$1</to>
</rule>

查看示例以获取更多...示例.

Answer 3

一个简单的基于BalusC回答步骤的JSF Url Prettyfier过滤器。过滤器将所有从/ ui路径开始的请求（假设您已经存储了所有xhtml文件）转发到同一路径，但是添加了xhtml后缀。

public class UrlPrettyfierFilter implements Filter {

    private static final String JSF_VIEW_ROOT_PATH = "/ui";

    private static final String JSF_VIEW_SUFFIX = ".xhtml";

    @Override
    public void destroy() {

    }

    @Override
    public void doFilter(ServletRequest request, ServletResponse response, FilterChain chain)
            throws IOException, ServletException {
        HttpServletRequest httpServletRequest = ((HttpServletRequest) request);
        String requestURI = httpServletRequest.getRequestURI();
        //Only process the paths starting with /ui, so as other requests get unprocessed. 
        //You can register the filter itself for /ui/* only, too
        if (requestURI.startsWith(JSF_VIEW_ROOT_PATH) 
                && !requestURI.contains(JSF_VIEW_SUFFIX)) {
            request.getRequestDispatcher(requestURI.concat(JSF_VIEW_SUFFIX))
                .forward(request,response);
        } else {
            chain.doFilter(httpServletRequest, response);
        }
    }

    @Override
    public void init(FilterConfig arg0) throws ServletException {

    }

}