麻煩映射一個URL到我的Java Servlet

Question

我需要通過輸入URL與下面的模式瀏覽器來從一個數據庫中的對象：

https://host.com/<something>?action=<actionName>&object=<node|interface>&selection=<list>&tenants=<list> 

哪裏「東西」是檢索PARAMATERS的方法和參數包括：行動，對象，選擇和租戶。

我該如何去建立web.xml文件中的模式以符合理解上述每個元素的方法？

<servlet> 
    <servlet-name>NewDynamicWebProject</servlet-name> 
    <servlet-class>com.test.package.NewDynamicWebProject</servlet-class> 
    </servlet> 

    <servlet-mapping> 
    <servlet-name>NewDynamicWebProject</servlet-name> 
    <url-pattern>/something/*</url-pattern> 
    </servlet-mapping> 

我的類：

@SuppressWarnings("serial") 
public class NewDynamicWebProject extends HttpServlet { 

    @Override 
    protected void doGet(HttpServletRequest req, HttpServletResponse resp) throws ServletException, IOException { 

     something(req, resp); 

    } 

    protected void something(HttpServletRequest req, HttpServletResponse resp) throws ServletException, IOException { 

     String actionName = req.getParameter("action"); 
     PrintWriter out = resp.getWriter(); 
     out.print("<div>" + actionName + "</div>"); 

    } 

} 

我怎麼能保證我的參數傳遞到我的網址是否有意義我的「東西」方法的範圍之內？

Answer 1

我放棄了「東西」類，只是做了以下內容：

@Override 
protected void doGet(HttpServletRequest req, HttpServletResponse resp) throws ServletException, IOException { 

    String actionName = req.getParameter("action"); 
    String objectType = req.getParameter("object"); 
    String selectionList = req.getParameter("selection"); 
    String tenantsList = req.getParameter("tenants"); 

    PrintWriter out = resp.getWriter(); 
    out.print("<div>" + actionName + " " + objectType + " " + selectionList + " " + tenantsList + "</div>"); 

}