grails的requestDispatcher的forward问题

java web开发关于请求的处理转发有两种方式，
  

• 一是redirect ，返回302状态和一个新Location给浏览器，让浏览器重新请求
•    二是requestDispatcher,可以用forward方法把请求转发给目标处理器。
grails 1.03现在的controller只提供了redirect方法，没有提供forward方法。
即使自己使用request.getRequestDispatcher("/controller/action").forward(request,response)方法或在Controller中返回ModelAndView("forward:/controller/action")，想让grails转发请求，grails也不支持，会得到404错误。

   web 2.4规范filter对请求的处理方式REQUEST(外部请求)、FORWARD(内部转发)、INCLUDE(包含) 和 ERROR(错误)，grails的URLMapping Filter只处理REQUEST,所以就处理不到forward的请求。
   解决方法：   [list]
  

1：修改grails源码和web.xml对filter配置，不过grails的jira中已经宣称在1.1版本中会在Controller中支持forward方法

   [*]2：看看grails的web.xml配置，所有的.dispatch结尾的请求会有GrailsDispatcherServlet来处理，在看看grails的源码

public class UrlMappingsFilter extends OncePerRequestFilter {
protected void doFilterInternal(HttpServletRequest request, HttpServletResponse response, FilterChain filterChain) throws ServletException, IOException {
...
if(viewName == null || viewName.endsWith(GSP_SUFFIX) || viewName.endsWith(JSP_SUFFIX)) {
                            String forwardUrl = WebUtils.forwardRequestForUrlMappingInfo(request, response, info);
                            if(LOG.isDebugEnabled()) {
                                LOG.debug("Matched URI ["+uri+"] to URL mapping ["+info+"], forwarding to ["+forwardUrl+"] with response ["+response.getClass()+"]");
                            }

                        }
}
}

public class WebUtils extends org.springframework.web.util.WebUtils {

		...
		
		public static String forwardRequestForUrlMappingInfo(HttpServletRequest request, HttpServletResponse response, UrlMappingInfo info) throws ServletException, IOException {
        return forwardRequestForUrlMappingInfo(request, response, info, Collections.EMPTY_MAP);
    }

    public static String forwardRequestForUrlMappingInfo(HttpServletRequest request, HttpServletResponse response, UrlMappingInfo info, Map model) throws ServletException, IOException {
        GrailsWebRequest webRequest = (GrailsWebRequest) RequestContextHolder.currentRequestAttributes();
        String forwardUrl = buildDispatchUrlForMapping(info);

        //populateParamsForMapping(info);
        RequestDispatcher dispatcher = request.getRequestDispatcher(forwardUrl);
        populateWebRequestWithInfo(webRequest, info);

        exposeForwardRequestAttributes(request);
        exposeRequestAttributes(request, model);
        dispatcher.forward(request, response);
        return forwardUrl;
    }
    
    public static String buildDispatchUrlForMapping(UrlMappingInfo info) {
        final StringBuffer forwardUrl = new StringBuffer();

        if (info.getViewName() != null) {
            String viewName = info.getViewName();
            forwardUrl.append(SLASH).append(viewName);
        }
        else {
        		// GRAILS_SERVLET_PATH="/grails" SLASH='/' GRAILS_DISPATCH_EXTENSION=".dispatch"
            forwardUrl.append(GrailsUrlPathHelper.GRAILS_SERVLET_PATH);
            forwardUrl.append(SLASH)
                              .append(info.getControllerName());

            if(!StringUtils.isBlank(info.getActionName())) {
                forwardUrl.append(SLASH)
                          .append(info.getActionName());
            }
            forwardUrl.append(GrailsUrlPathHelper.GRAILS_DISPATCH_EXTENSION);
        }

        return forwardUrl.toString();
    }
}

按照grails处理规则，只要forward的url为/grails/controller/action.dispatch，grails就可以处理了。
  [/list]
   所以暂时的解决方法就是返回一个ModelAndView(forwardurl),forwardurl的格式为"forward:/grails/${controllerName}/${actionName}.dispatch"