Question

运行项目时，我收到此“此URL不支持HTTP方法POST”错误。有趣的是，两天前它运行得非常好。我对我的代码进行了一些更改后，然后恢复了我的原始代码，并给了我这个错误。你能帮我吗？

这是我的index.html：

<form method="post" action="login.do">
<div>
<table>
            <tr><td>Username: </td><td><input type="text" name="e_name"/>
            </td>  </tr>
            <tr><td> Password: </td><td><input type="password" name="e_pass"/>
            </td>  </tr>
            <tr><td></td><td><input type="submit" name ="e_submit" value="Submit"/>

这是我的登录servlet：

public class Login extends HttpServlet {

/**
 * Processes requests for both HTTP
 * <code>GET</code> and
 * <code>POST</code> methods.
 *
 * @param request servlet request
 * @param response servlet response
 * @throws ServletException if a servlet-specific error occurs
 * @throws IOException if an I/O error occurs
 */
protected void processRequest(HttpServletRequest request, HttpServletResponse response)
        throws ServletException, IOException, SQLException {
    response.setContentType("text/html;charset=UTF-8");
    PrintWriter out = response.getWriter();
    try {
        /*
         * TODO output your page here. You may use following sample code.
         */
        int status;
        String submit = request.getParameter("e_submit");
        String submit2 = request.getParameter("a_submit");
        out.println("Here1");
        String e_name = request.getParameter("e_name");
        String e_password = request.getParameter("e_pass");
        String a_name = request.getParameter("a_name");
        String a_password = request.getParameter("a_pass");
        out.println(e_name+e_password+a_name+a_password);
        Author author = new Author(a_name,a_password);  
        Editor editor = new Editor(e_name,e_password);

      // If it is an AUTHOR login:

       if(submit==null){
       status = author.login(author);
       out.println("Author Login");
       //Incorrect login details
       if(status==0) {
           out.println("Incorrect");

       RequestDispatcher view = request.getRequestDispatcher("index_F.html");
       view.forward(request, response);

       }
       //Correct login details --- AUTHOR

       else {

              out.println("Correct login details");
              HttpSession session = request.getSession();    
              session.setAttribute(a_name, "a_name");

              RequestDispatcher view = request.getRequestDispatcher("index_S.jsp"); 
              view.forward(request, response);
            }

       }

       //If it is an EDITOR login

       else if (submit2==null){

           status = editor.login(editor);

           //Incorrect login details

           if(status==0) {

       RequestDispatcher view = request.getRequestDispatcher("index_F.html");
       view.forward(request, response);
            }

           //Correct login details --- EDITOR

           else {
               out.println("correct");
               HttpSession session = request.getSession();    
       session.setAttribute(e_name, "e_name");
       session.setAttribute(e_password, "e_pass");
               RequestDispatcher view   = request.getRequestDispatcher("index_S_1.html"); 
               view.forward(request, response);

            }           }


        out.println("</body>");
        out.println("</html>");



    } finally {            
        out.close();
    }
}

@Override
protected void doPost(HttpServletRequest req, HttpServletResponse resp) throws ServletException, IOException {
    super.doPost(req, resp);
}

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

我的web.xml看起来像这样：

<?xml version="1.0" encoding="UTF-8"?>
<servlet>
    <servlet-name>action</servlet-name>
    <servlet-class>org.apache.struts.action.ActionServlet</servlet-class>
    <init-param>
        <param-name>config</param-name>
        <param-value>/WEB-INF/struts-config.xml</param-value>
    </init-param>
    <init-param>
        <param-name>debug</param-name>
        <param-value>2</param-value>
    </init-param>
    <init-param>
        <param-name>detail</param-name>
        <param-value>2</param-value>
    </init-param>
    <load-on-startup>2</load-on-startup>
</servlet>
<servlet>
    <servlet-name>Login</servlet-name>
    <servlet-class>controller.Login</servlet-class>
</servlet>
<servlet-mapping>
    <servlet-name>action</servlet-name>
    <url-pattern>*.do</url-pattern>
</servlet-mapping>
<servlet-mapping>
    <servlet-name>Login</servlet-name>

    <url-pattern>/login.do</url-pattern>
</servlet-mapping>
<session-config>
    <session-timeout>
        30
    </session-timeout>
</session-config>
<welcome-file-list>
    <welcome-file>index.html</welcome-file>
</welcome-file-list>

我使用Glassfish v3服务器 - 让我知道您需要知道的任何其他内容

Answer 1

那是因为在您的doGet()和doPost()方法中，您正在调用它的super方法。而是在上面提到的相应方法中调用processRequest()。

super.doGet()和super.doPost()方法默认返回HTTP 405，因此您不会调用超类doGet()和doPost()。

Answer 2

为什么代码中有processRequest方法？谁会称这种方法？

通过致电super.doGet()或super.doPost()，您无法达到该方法你需要明确地调用那个方法。

@Override
protected void doPost(HttpServletRequest req, HttpServletResponse resp) throws ServletException, IOException {
    processRequest(req,resp)
}

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

修改

这样做

response.sendRedirect("index_F.html"); return;

而不是

RequestDispatcher view = request.getRequestDispatcher("index_F.html"); view.forward(request, response);

Answer 3

在doPost()内，你必须致电processRequest()。

Answer 4

您在没有实际实施（使用doGet()）的情况下调用doPost()和super方法。

HttpServlet基本上遵循模板方法模式，其中所有未重写的HTTP方法都返回HTTP 405错误“Method not supported”。当您覆盖此类方法时，不应调用super方法，否则您仍会收到HTTP 405错误。你的doPost（）方法也是如此。

在上述相应方法中调用processRequest(req,resp)。

编辑：

其次，

不要使用调度程序将请求转发给HTML。如果你只想显示html，请使用重定向。

response.sendRedirect("index_F.html");
return;

此外，最好在执行注销时使用redirect或发回无效凭据。

HTTP状态405 - 此URL不支持HTTP方法POST

4 个答案: