我需要将请求转发到REST服务,并编写了以下代码:
private void performTask(HttpServletRequest request, HttpServletResponse response) throws ServletException,
IOException {
ServletContext portalContext = this.getServletContext();
ServletContext restService = portalContext.getContext("/restService");
RequestDispatcher dispatcher = restService.getRequestDispatcher("/resources/*");
dispatcher.forward(request, response);
}服务"/restService“也有一个servlet来处理请求,它的定义如下:
<servlet>
<servlet-name>Jersey REST Service</servlet-name>
<servlet-class>org.glassfish.jersey.servlet.ServletContainer</servlet-class>
<init-param>
<param-name>javax.ws.rs.Application</param-name>
<param-value>com.service.RESTApplication</param-value>
</init-param>
<init-param>
<param-name>jersey.config.server.provider.packages</param-name>
<param-value>com.rest.resources</param-value>
</init-param>
<load-on-startup>1</load-on-startup>
</servlet>
<servlet-mapping>
<servlet-name>Jersey REST Service</servlet-name>
<url-pattern>/resources/*</url-pattern>
</servlet-mapping>类RESTApplication具有以下代码:
public RESTApplication() {
register(WadlFeature.class);
register(JacksonFeature.class);
register(MultiPartFeature.class);
property(CommonProperties.FEATURE_AUTO_DISCOVERY_DISABLE, true);
}在"com.rest.resources“中有几个类映射到请求附带的路径(/resources/*)。,但是当请求被"getRequestDispatcher“转发时,泽西会抛出一个404错误。似乎泽西无法将我的请求映射到正确的servlet。如果我使用Postman粘贴相同的URL,请求将被正常处理。调度员的检查是:

为了验证这个问题是否与泽西有关,我创建了一个从HttpServlet扩展的Servlet,它没有提交给泽西:
<servlet>
<servlet-name>HelloServlet</servlet-name>
<servlet-class>com.rest.resources.helloResource</servlet-class>
<load-on-startup>1</load-on-startup>
</servlet> 在此之后,我执行了以下代码:
private void performTask(HttpServletRequest request, HttpServletResponse response) throws ServletException,
IOException {
ServletContext portalContext = this.getServletContext();
ServletContext restService = portalContext.getContext("/restService");
RequestDispatcher dispatcher = restService.getRequestDispatcher("/hello");
dispatcher.forward(request, response);
}令人惊讶的是,这个转发实现了helloResource类,该类被放置到"com.rest.resources“中。
有人遇到过类似的情况吗?"getRequestDispacther“如何找到和转发存在于”/resources/“中的servlet?
使用的版本:Jersey2.6Servlet2.5JBos4.2.3-GA
发布于 2017-08-30 13:43:04
看来,泽西希望"getRequestDispatcher“实现必须处理的URL,而不是将其委托给映射的servlet。
这条线解决了这个问题:
datasulRest.getRequestDispatcher("/resources/“+request.getPathInfo();
在此之后,该请求由放置在"com.rest.resources“中的资源处理。
https://stackoverflow.com/questions/45939660
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