Question

我的网站遇到问题，我试图让用户插入用户名和密码登录主页，但它显示我的错误

HTTP状态404-请求的资源（）不可用

似乎html页面没有从servlet获得任何结果。我不确定，但有人可以帮助我并告诉我我错过了什么吗？

这是我的servlet：

import java.util.logging.Level;
import java.util.logging.Logger;
import javax.servlet.*;
import javax.servlet.http.*;
import java.io.*;
import java.sql.*;

public class enter extends HttpServlet {

  private Connection myCon;
  private PreparedStatement myStmt;

    @Override
  public void init() {

System.out.println("init");
    try {

      Class.forName("com.mysql.jdbc.Driver").newInstance();
    }
    catch (Exception e) {
    }

       try {
        System.out.println("connecting");
      myCon = DriverManager.getConnection ("jdbc:mysql://localhost/auk","root","password");

    }
    catch (SQLException e) {
    }
  } 

    protected void processRequest(HttpServletRequest request, HttpServletResponse response)
    throws ServletException, IOException, SQLException {
    response.setContentType("text/html;charset=UTF-8");
    ResultSet result;
    PrintWriter out = response.getWriter();
    String query, colName, dat, user;
    int numCols, index;
    ResultSetMetaData resultMd;

     query = "Select Pass from user where Username=?";

    try {
      myStmt = myCon.prepareStatement(query);
    }
    catch (Exception e) {
    }
         user=request.getParameter("user");
     myStmt.setString(1, user);   

     result=myStmt.executeQuery();
     String userpass=result.getString(1);

     out.println ("userpass selected from the table is = "+userpass);

    response.setContentType("text/html");

    out.println("<html>");
    out.println("<head><title>JDBCServlet</title></head>");
    out.println("<body>");
    out.print("<p><b>The query is: </b>" + query + "</p>");

    try {

      result = myStmt.executeQuery(query);

      resultMd = result.getMetaData();
      numCols = resultMd.getColumnCount();

      out.println("<table border>");
      out.println("<caption> <b> Query Results </b> </caption>");
      out.println("<tr>");

     out.print("<th> Welcome " + user + " </th>");
      for (index = 1; index <= numCols; index++) {
        colName = resultMd.getColumnLabel(index);
        out.print("<th>" + colName + "</th>");
      }
        out.println("</tr>");

        while (result.next()) {
        out.println("<tr>");

        for (index = 0; index < numCols; index++) {
        dat = result.getString(index + 1);
        out.println("<td>" + dat + "</td>");
      } 

      out.println("</tr>");
    } 
    out.println("</table>");
  } 
    catch (Exception e) {
      e.printStackTrace();
    } 
    out.println("</body></html>");

    }     

    @Override
    protected void doGet(HttpServletRequest request, HttpServletResponse response)
    throws ServletException, IOException {
        try {
            processRequest(request, response);
        } catch (SQLException ex) {
            Logger.getLogger(enter.class.getName()).log(Level.SEVERE, null, ex);
        }
    } 

    @Override
    protected void doPost(HttpServletRequest request, HttpServletResponse response)
    throws ServletException, IOException {
        try {
            processRequest(request, response);
        } catch (SQLException ex) {
            Logger.getLogger(enter.class.getName()).log(Level.SEVERE, null, ex);
        }
    }

    @Override
    public String getServletInfo() {
        return "Short description";
    }
  }

Answer 1

该错误仅表示没有任何内容正在侦听URL。

因此，URL是完全错误的（区分大小写！），或者Servlet根本没有侦听该URL，或者Servlet无法启动。根据迄今给定的信息很难说出来。您需要验证浏览器地址栏中的URL，web.xml中的servlet映射以及服务器启动日志。

对于具体问题

无关，将像Connection这样昂贵的数据库资源分配为servlet的实例变量是非常糟糕的主意。您应该在同一方法中以尽可能短的范围打开和关闭这些资源。

登录servlet页面的问题

1 个答案: