Question

我在一个简单的servlet链接示例中看到一个奇怪的问题，我正在尝试：

Servlet 1：

public class gatewayservlet extends HttpServlet {

  public void doPost(HttpServletRequest request , 
    HttpServletResponse response)
    throws ServletException , IOException {
        doGet(request,response);
  }

  public void doGet(HttpServletRequest request , 
    HttpServletResponse response)
    throws ServletException , IOException {

    response.setContentType("text/plain");

    PrintWriter out = response.getWriter();

    name = request.getParameter("name");

    RequestDispatcher rd = getServletConfig().getServletContext().getRequestDispatcher("/justServlets/secondservlet");

    if(name!=null) {
      request.setAttribute("UserName",name);
      rd.forward(request , response);
      // Forward the value to another Secondservlet
    } else {
      response.sendError(response.SC_BAD_REQUEST, 
        "UserName Required");
    }

  }

}

Servlet 2：

public class secondservlet extends HttpServlet {

  public void doGet(HttpServletRequest request , 
    HttpServletResponse response)
    throws ServletException , IOException {

    response.setContentType("text/plain");

    PrintWriter out = response.getWriter();

    String UserName = (String)request.getAttribute("UserName");

    out.println("The UserName is "+ UserName);

  }


  public void doPost(HttpServletRequest request , 
    HttpServletResponse response)
    throws ServletException , IOException {
        doGet(request,response);
  }

}

以及调用表单：

<html>
<body>
<FORM ACTION="/justServlets/gateway" METHOD=POST>
<P>Please Fill the Registration Form</p><br>
Enter Your Name<input type="text" name="name"><br>
<input type="submit" value="send">
</FORM>
</body>
</html>

'POST'给出405（不允许方法）错误。但是，调用第一个servlet为.... / justServlets / gateway？name =苏格拉底有效。怎么回事？

Answer 1

假设/justServlets是您的网络应用的上下文路径，转发的代码应该使用/secondservlet而不是/justServlets/secondservlet，因为正如the javadoc所说：

路径名必须以/开头，并且相对于当前上下文根被解释为。

（强调我的）

实际上，您转发到/justServlets/justServlets/secondservlet，这可能不存在。

Answer 2

当您转发的servlet缺少与最初处理的方法相同的处理程序（在本例中为POST）时，通常会发生这种情况。确保servlet2确实处理了doPost，并确实转发到servlet2。

Servlet链接 - 简单的例子

2 个答案: