Jersey URL forwarding | 易学教程

问题

Inside a Jersey REST method I would like to forward to an another website. How can I achieve that?

@Path("/")
public class News {

    @GET
    @Produces(MediaType.TEXT_HTML)
    @Path("go/{news_id}")
    public String getForwardNews(
        @PathParam("news_id") String id) throws Exception {

        //how can I make here a forward to "http://somesite.com/news/id" (not redirect)?

        return "";
    }
}

EDIT:

I'm getting the No thread local value in scope for proxy of class $Proxy78 error when trying to do something like this:

@Context
HttpServletRequest request;
@Context
HttpServletResponse response;
@Context
ServletContext context;

...

RequestDispatcher dispatcher =  context.getRequestDispatcher("url");
dispatcher.forward(request, response);

回答1:

public Response foo()
{
    URI uri = UriBuilder.fromUri(<ur url> ).build();
    return Response.seeOther( uri ).build();
}

I used above code in my application and it works.

回答2:

Try this, it worked for me:

public String getForwardNews(

@Context final HttpServletRequest request,

@Context final HttpServletResponse response) throws Exception

{

System.out.println("CMSredirecting... ");

response.sendRedirect("YourUrlSite");

return "";

}

回答3:

I can't test it right now. But why not...

Step 1. Get access to HttpServletResponse. To do it declare in your service something like:

@Context
HttpServletResponse _currentResponse;

Step 2. make redirect

...
_currentResponse.sendRedirect(redirect2Url);

EDIT

Well, to call forward method you need get to ServletContext. It is can be resolved the same way as response:

@javax.ws.rs.core.Context 
ServletContext _context;

Now _context.forward is available

来源：https://stackoverflow.com/questions/8242719/jersey-url-forwarding

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