Question

是否可以在发送请求时更改网址。

这是我的代码

public void doGet(HttpServletRequest req, HttpServletResponse res) throws IOException, ServletException
{

 List<HomePageServicesDescription> data= HomePageServicesDescriptionDB.showHomePageServicesDescription();
 req.setAttribute("description", data);

 req.getRequestDispatcher("index.jsp").forward(req,res);

 }

因此，当在Web浏览器中看到它时，它会给出servlet的url = http://localhost:8888/url-mapping。但我想要url = http://localhost:8888/index.jsp。怎么可能。

Answer 1

您应该HttpServletResponse.sendRedirect()而不是RequestDisaptcher.forward()。您要发送的任何参数都可以作为查询参数发送。

public void doGet(HttpServletRequest req, HttpServletResponse res) 
        throws IOException, ServletException
{

   List<HomePageServicesDescription> data =
             HomePageServicesDescriptionDB.showHomePageServicesDescription();
    req.setAttribute("description", data);

    res.sendRedirect("index.jsp?description="+data);

 }

Answer 2

我得到了答案

public void doGet(HttpServletRequest req, HttpServletResponse res) 
    throws IOException, ServletException
{

    List<HomePageServicesDescription> data = HomePageServicesDescriptionDB.showHomePageServicesDescription();
    req.getSession().setAttribute("description", data);

    res.sendRedirect("index.jsp");

}

在index.jsp

中

List<HomePageServicesDescription> data= (List<HomePageServicesDescription>) session.getAttribute("description");

完美的工作

可以更改请求调度程序中的URL

2 个答案: