Question

我们显然坚持使用登录程序，因为它没有读取用户名和密码。它运行，即使用户名和密码不正确。

实验工作代码（即使登录详细信息不正确也登录）：

JButton btnLogin = new JButton("Log-In");
    btnLogin.addActionListener(new ActionListener() {
        public void actionPerformed(ActionEvent arg0) {



            Connection conn;
            java.sql.PreparedStatement ps;

              try{
                  conn = DriverManager.getConnection("jdbc:mysql://localhost:3306/tupaness","ivanaldwin","frankscanteen");
                  Statement mystmt = conn.createStatement();
                  ps = conn.prepareStatement("SELECT * FROM 'admin' WHERE 'uname' = ? AND 'pass' = ?");


                  String user = uname.getText();   // Collecting the input
                   char[] password = pass.getPassword(); // Collecting the input
                   String pwd = String.valueOf(pass);  // converting from array to string
                   if(validate_login(user,pwd)){

                 /* ps.setString(1, uname.getText());
                  ps.setString(2, String.valueOf(pass.getPassword()));
                  ResultSet result = ps.executeQuery();



                  if(result.next()){
                      JOptionPane.showMessageDialog(null, "Login Success");

                      TupanessHome th = new TupanessHome();
                  }
                  */
                   }
                }catch(Exception e) {
                      JOptionPane.showMessageDialog(null, "An Error has been Detected: \n" + e);
                }TupanessHome th = new TupanessHome();
        }

        private boolean validate_login(String user, String pwd) {

            try{           
                   Class.forName("com.mysql.jdbc.Driver");  // MySQL database connection
                   Connection conn = DriverManager.getConnection("jdbc:mysql://localhost:3306/tupaness","ivanaldwin","frankscanteen");     
                   PreparedStatement pst = conn.prepareStatement("Select * from admin where uname=? and pass=?");
                   pst.setString(1, user); 
                   pst.setString(2, pwd);
                   ResultSet rs = pst.executeQuery();                        
                   if(rs.next()){            
                       return true;                
                   }
                   else
                       return false;            
               }
               catch(Exception e){
                   e.printStackTrace();
                   return false;
               }       



        }
    });
    btnLogin.setBounds(65, 117, 89, 23);
    login.getContentPane().add(btnLogin);

应该是代码（不能正常工作，说我们有某种语法错误）

    JButton btnLogin = new JButton("Log-In");
    btnLogin.addActionListener(new ActionListener() {
        public void actionPerformed(ActionEvent arg0) {
            Connection conn;
            java.sql.PreparedStatement ps;

              try{
                  conn = DriverManager.getConnection("jdbc:mysql://localhost:3306/tupaness","ivanaldwin","frankscanteen");
                  ps = conn.prepareStatement("SELECT * FROM 'admin' WHERE 'uname' = ? AND 'pass' = ?");
                  ps.setString(1, uname.getText());
                  ps.setString(2, String.valueOf(pass.getPassword()));
                  ResultSet result = ps.executeQuery();

                  if(result.next()){
                      JOptionPane.showMessageDialog(null, "Login Success");

                      TupanessHome th = new TupanessHome();
                  }

                }catch(Exception e) {
                      JOptionPane.showMessageDialog(null, "An Error has been Detected: \n" + e);
                }
        }
    });
    btnLogin.setBounds(65, 117, 89, 23);
    login.getContentPane().add(btnLogin);

修改

我们设法修复了列中的错误，但它仍然无法登录

  ps = conn.prepareStatement("SELECT * FROM admin WHERE 'uname' = ? AND 'pass' = ?");

（声明来自＆＃39;应该是代码＆＃39;）

Answer 1

"SELECT * FROM 'admin' WHERE 'uname' = ? AND 'pass' = ?"

应该是

"SELECT * FROM `admin` WHERE `uname` = ? AND `pass` = ?"

甚至

"SELECT * FROM admin WHERE uname = ? AND pass = ?"

Java SQL - 无法正确读取UserInput（用户名和密码）

1 个答案: