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i have tried with all these things but giving url like
/get.do. i have two forwaed pages success and failure.if success i want url like ..../projectname/success.jsp. how can i get this one
 
i have used the following code but none of working.
 
public class LoginAction extends Action {
   public ActionForward execute(ActionMapping mapping, ActionForm form,
            HttpServletRequest request, HttpServletResponse response)
            throws Exception {
       String status;
       String msg=null;
      String originalURI=(String)request.getAttribute(org.apache.struts.Globals.ORIGINAL_URI_KEY);
      
       System.out.println("originalurl========"+originalURI);
       Loginform lf=(Loginform)form;
       String name=lf.getName();
       String pwd=lf.getPwd();
       if(name.equals("swapna") && pwd.equals("swappna"))
       {
          status="success"; 
       }
       else
       {
           status="Failure";
       }
      //System.out.println("hfjkfhkfhef===="+ mapping.getPath()); 
       String url=request.getAttribute("javax.servlet.forward.request_uri");
       String url1=request.getAttribute("javax.servlet.forward.query_string")
       System.out.println("path========-----"+url);
       return mapping.findForward(status);
   }
}
 

 
is there any way to get original url.
Posted 24-May-13 23:56pm

1 solution

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Solution 1

Hello,
 
You can obtain the actual URL by using code similar to one shown below.
ActionForward af = mapping.findForward(status);
System.out.println("path========-----".concat(af.getPath()));
Regards,
  Permalink  

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