当我跑我的项目，我得到这个“HTTP POST方法不受此URL支持”错误。 有趣的是，它运行完全正常，前两天。 之后，我做我的代码进行一些更改，但随后恢复了我的原代码和它给我这个错误。 请你帮助我好吗？
这里是我的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"/>
这里是我的登录的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);
}}
而我的web.xml看起来是这样的：
<?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>
我用Glassfish的第三版服务器 - 让我知道其他任何你需要知道的
