Question

在泽西岛,我想向另一个网站转播。我如何能够做到这一点?

@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 "";
    }
}

http://www.un.org。

页: 1 替代类别(Proxy78/code)的本地价值在试图做这样的事情时有错误:

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

...

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

Answer 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.

Answer 2

为此,它为我工作:

public String getForwardNews(

@Context final HttpServletRequest request,

@Context final HttpServletResponse response) throws Exception

{

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

response.sendRedirect("YourUrlSite");

return "";

}

Answer 3

我现在可以试一下。但原因并非如此。

步骤1. 进入HttpServletResponse。声明如下:

@Context
HttpServletResponse _currentResponse;

第2步:调整方向

...
_currentResponse.sendRedirect(redirect2Url);

http://www.ohchr.org。

当然,为了向前推进,你需要找到“SerletContext”。解决方式与应对方式相同:

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

Now _context.forward is available

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