Question

当我使用方法响应并且我没有任何关于如何修复它的线索时，我收到上述消息，我有java servlet 3.1这似乎是我能在其他地方找到的唯一答案解决它。

import java.io.*;
import javax.servlet.*;
import javax.servlet.http.*;
import java.sql.*;

public class login extends HttpServlet {

protected void doPost(HttpServletRequest request, HttpServlet response) throws ServletException, IOException {
    response.setContentType("text/html;charset=UTF-8");
    PrintWriter out = response.getWriter();

    String username = request.getParameter("username");
    String password = request.getParameter("password");

    if(Validate.checkUser(username, password))
    {
        RequestDispatcher rs = request.getRequestDispatcher("Welcome");
        rs.forward(request, response);
    }
    else
    {
        out.println("Username or Password incorrect");
        RequestDispatcher rs = request.getRequestDispatcher("login.html");
        rs.include(request, response);
    }
}

}

Answer 1

检查你发布的方法 HttpServlet有doPost方法接受HttpServleRequest和HttpServletResponse参数

protected void doPost(HttpServletRequest request, HttpServlet response) throws ServletException, IOException {

到

protected void doPost(HttpServletRequest request, HttpServletResponse response) throws     ServletException, IOException {

Java中类型HttpServlet错误的方法未定义

1 个答案: