从servlet发生异常时如何重定向到错误页面？

Question

I am writing a servlet, in that if any exception occurs i donэt want to display exception/error message on browser, so I will redirect to my customized error page. So I have done like this:

protected void doPost(HttpServletRequest request,
        HttpServletResponse response) throws ServletException, IOException {
try{
    //Here is all code stuff
}catch(Exception e){

  request.getRequestDispatcher("/ErrorPage.jsp").forward(request, response);
  e1.printStackTrace();

}

Is this the correct way, if I am wrong please correct me and if there is any better mechanism please tell me.

Answer 1

Only way to handle it in a generic way is to use web.xml like below:

<error-page>
  <exception-type>java.lang.Throwable</exception-type>
  <location>/ErrorHandler</location>
</error-page>

The servlet is thrown ServletException and IOException but if you want to handle runtime exceptions and all other exceptions in a single exception handler, you can provide exception-type as Throwable . You can use multiple error-page entries that will handle different type of exceptions and have different handlers.

Example:

@WebServlet("/ErrorHandler")
public class ErrorHandler extends HttpServlet {
    private static final long serialVersionUID = 1L;

    protected void doGet(HttpServletRequest request,
            HttpServletResponse response) throws ServletException, IOException {
        processError(request, response);
    }

    protected void doPost(HttpServletRequest request,
            HttpServletResponse response) throws ServletException, IOException {
        processError(request, response);
    }
    private void processError(HttpServletRequest request,
            HttpServletResponse response) throws IOException {
        //customize error message
        Throwable throwable = (Throwable) request
                .getAttribute("javax.servlet.error.exception");
        Integer statusCode = (Integer) request
                .getAttribute("javax.servlet.error.status_code");
        String servletName = (String) request
                .getAttribute("javax.servlet.error.servlet_name");
        if (servletName == null) {
            servletName = "Unknown";
        }
        String requestUri = (String) request
                .getAttribute("javax.servlet.error.request_uri");
        if (requestUri == null) {
            requestUri = "Unknown";
        }    
        request.setAttribute("error", "Servlet " + servletName + 
          " has thrown an exception " + throwable.getClass().getName() +
          " : " + throwable.getMessage());    
        request.getRequestDispatcher("/ErrorPage.jsp").forward(request, response);
    }
}

Answer 2

One way to handle it in a generic way is to use web.xml like below:

<error-page>
    <exception-type>java.io.IOException</exception-type >
    <location>/ErrorHandler</location>
</error-page>

I have just included IO exception but you might have say SQLException you could very well add another error-page and another location for the same. Similarly you could say java.lang.Exception type and one handler handling everything.

Answer 3

In some method, you would have the following:

try {
  // something
} catch (Exception e) {
  sendErrorRedirect(req, res, "/errorpage.jsp", e);
}

// then....  sendErrorRedirect looks like this:
  protected void sendErrorRedirect(HttpServletRequest request, HttpServletResponse response, String errorPageURL, Throwable e) {
      try {
            request.setAttribute ("javax.servlet.jsp.jspException", e);
            getServletConfig().getServletContext().getRequestDispatcher(errorPageURL).forward(request, response);
      } catch (Exception ex) {
            putError("serXXXXX.sendErrorRedirect ", ex);
      }
  }