Question

我正在使用jsp调用servlet

//My servlet code is:
public void doGet(HttpServletRequest request, HttpServletResponse response)
       {
           String template="test";   
           abcViewBean punchOutCan = new abcViewBean();
           punchOutCan.setPunchOutCanonicalRes(template);
           try {
            request.getRequestDispatcher("/PunchOutCanonicalError.jsp").forward(request,response);
        } catch (ServletException e) {
            // TODO Auto-generated catch block
            e.printStackTrace();
        } catch (IOException e) {
            // TODO Auto-generated catch block
            e.printStackTrace();
        }
       }

我的JSP代码是：

<jsp:include page="/PunchOutCanonicalServlet" flush="true"/>  
<c:out value="${punchOutCan.punchOutCanonicalRes}" />

请建议，如何摆脱这个。

Answer 1

从servlet的doGet中排除（删除）此语句，因为您正在JSP中导入响应。

request.getRequestDispatcher("/PunchOutCanonicalError.jsp")
    .forward(request,response);

doGet必须是：

@Override
public void doGet(HttpServletRequest request, 
                  HttpServletResponse response)
                    throws ServletException,IOException{
       String template="test";   
       abcViewBean punchOutCan = new abcViewBean();
       punchOutCan.setPunchOutCanonicalRes(template);
       //You can push the bean object into request via setAttribute
       //e.g
       //request.setAttribute("punchOutCan",punchOutCan);
}

和JSP Code，

<jsp:include page="/PunchOutCanonicalServlet" flush="true"/>  
<c:out value="${punchOutCan.punchOutCanonicalRes}" />

无法转发。响应已经提交。错误

1 个答案: