Java servlet redirect between two different jersey REST services -

we have existing jersey rest service in our application(url: /rest/*). sample url looks like: http://xxx:8080/app/rest/company/getdata

based on property need redirect rest call context(url: /newrest/*): sample url like: http://xxx:8080/app/newrest/company/getdata

so, added servlet mapping in web.xml.

so, web.xml looks below snippet.

<servlet>     <servlet-name>servletadaptor</servlet-name>     <servlet-class>com.sun.jersey.spi.container.servlet.servletcontainer</servlet-class>     <init-param>         <param-name>com.sun.jersey.config.property.packages</param-name>         <param-value>xxx.httpservice</param-value>     </init-param>     <load-on-startup>1</load-on-startup> </servlet> <servlet-mapping>     <servlet-name>servletadaptor</servlet-name>     <url-pattern>/rest/*</url-pattern> </servlet-mapping>   <filter>     <filter-name>securerestfilter</filter-name>     <filter-class>xxx.restsecurityfilter</filter-class> </filter> <filter-mapping>     <filter-name>securerestfilter</filter-name>     <url-pattern>/rest/*</url-pattern> </filter-mapping>  <servlet>     <servlet-name>newadaptor</servlet-name>     <servlet-class>com.sun.jersey.spi.container.servlet.servletcontainer</servlet-class>     <init-param>         <param-name>com.sun.jersey.config.property.packages</param-name>         <param-value>xxx.newhttpservice</param-value>     </init-param>     <load-on-startup>1</load-on-startup> </servlet> <servlet-mapping>     <servlet-name>newadaptor</servlet-name>     <url-pattern>/newrest/*</url-pattern> </servlet-mapping> 

i have written filter(to check property on servletadaptor servlet , need redirect newadaptor servlet, if needed. also, final response should sent servlet adaptor servlet using response newadaptor servlet, if necessary(based on property).

i need directions solving issue. please help.

the dofilter method in filter class looks like:

@override public void dofilter(servletrequest req, servletresponse res, filterchain chain) throws servletexception, ioexception {          httpservletrequest request = (httpservletrequest) req;         string productmode = "b";//hardcoded example         string url = request.getrequesturl().tostring();     if(productmode.equals("a")){         chain.dofilter(req, res);     }else{         url=url.replace("rest", "newrest");         req.getrequestdispatcher(url).forward(req, res);     }  } 

but requestdispatcher not seem forward request new url.

from understand trying create new services or integrate newer one. ideally can version url's support backward compatibilty in future.e.g. http://{xxx}:8080/app/rest/v1_0/company/getdata,http://{xxx}:8080/app/rest/v2_0/company/getdata etc.

why don't wrap new api call in existing api call? consumer talks api, internally invoke newer api , consume response of it.then either return new object or wrap object around it.

you need re-look @ strategy trying adopt!