Java JDBC swing login form error

Question

I am currently working on a simple JDBC project and I am stuck at the login form

 private void jButton1ActionPerformed(java.awt.event.ActionEvent evt) {                                         
    try
{
    Class.forName("com.mysql.jdbc.Driver");
    con=DriverManager.getConnection("jdbc:mysql://localhost/company","root","redhat");
    stmt=con.createStatement(); 
    String nm=jTextField1.getText();
    char[] pass=jPasswordField1.getPassword();
    String pw=Arrays.toString(pass);
    String sql="select * from login where uname='"+nm+"' and pass='"+pw+"'";
    rs=stmt.executeQuery(sql);
    if(rs.next())
     {
      new MainPage().setVisible(true);
      this.setVisible(false);
     }
    else
     {
      JOptionPane.showMessageDialog(this, "Wrong User name or password");
      jTextField1.setText("");
      jPasswordField1.setText("");

     }
}
catch(Exception e)
{
  System.out.println(e);
}
}              `

When this button is clicked a new page that I have created is to be shown ,but it shows "Wrong User name or password" message dialog. What are my mistakes? My database is in mysql.

try to print both values nm+" and "+pw and you will see the difference — Pavneet_Singh
– Pavneet_Singh, Commented Dec 12, 2016 at 8:58

Andreas · Accepted Answer · 2016-12-12 09:07:43Z

Error 1: Arrays.toString(char[]) will return an array representation, e.g. if the password in jPasswordField1 is password, the result is this string: [p, a, s, s, w, o, r, d]

Fix 1: Use new String(char[]) instead.

Error 2: Using string concatenation to build a SQL statement. This will make your code susceptible to SQL Injection attacks, where hackers can steal your data and delete your tables.

Fix 2: Use a PreparedStatement.

Error 3: Not releasing resources. This will lead to memory leaks.

Fix 3: Use try-with-resources.

Result:

private void jButton1ActionPerformed(java.awt.event.ActionEvent evt) {                                         
    try {
        boolean loginOk;
        Class.forName("com.mysql.jdbc.Driver");
        try (Connection con = DriverManager.getConnection("jdbc:mysql://localhost/company","root","redhat")) {
            String sql = "select * from login where uname=? and pass=?";
            try (PreparedStatement stmt = con.prepareStatement(sql)) {
                stmt.setString(1, jTextField1.getText());
                stmt.setString(2, new String(jPasswordField1.getPassword()));
                try (ResultSet rs = stmt.executeQuery(sql)) {
                    loginOk = rs.next();
                }
            }
        }
        if (loginOk) {
            new MainPage().setVisible(true);
            this.setVisible(false);
        } else {
            JOptionPane.showMessageDialog(this, "Wrong User name or password");
            jTextField1.setText("");
            jPasswordField1.setText("");
        }
    } catch(Exception e) {
        System.out.println(e);
    }
}

Shivakumar ss · Accepted Answer · 2016-12-12 09:03:39Z

0

Arrays.ToString Returns a string representation of the contents of the specified array. The string representation consists of a list of the array's elements.

In your case Instead of

String pw = Arrays.toString(pass);

You should be using to get the password in String representation.

String pw = String.copyValueOf(pass);

answered Dec 12, 2016 at 9:03

Shivakumar ss

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Comments

Riaan Nel · Accepted Answer · 2016-12-12 09:02:40Z

0

That would seem to indicate that your query returns an empty result set. What happens if you run it against the DB directly?

Also, Arrays.toString(...) will give you a String representation of the array. You probably want the contents of the array in a String - there's a difference. Try using new String(pass) instead.

answered Dec 12, 2016 at 9:02

Riaan Nel

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Romeo · Accepted Answer · 2016-12-12 09:02:40Z

0

I think "Wrong User name or password" means rs is null or rs'size is 0. You should check the name and password.

answered Dec 12, 2016 at 9:02

Romeo

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