java sql exception occures repeatedly [duplicate]

Question

public void Deposite() throws Exception

    {
            try
            {
                Class.forName("com.mysql.jdbc.Driver");

                String url = "jdbc:mysql://localhost:3306/bank";

                Connection con = DriverManager.getConnection(url,"root","admin");

                BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
                System.out.print("Enter your A/c no. : " );
                acNo = Integer.parseInt(br.readLine());

                String sql = "SELECT Name,Ac_No,Balance FROM CUSTOMER WHERE Ac_No=?";
                PreparedStatement ps = con.prepareStatement(sql);
                ps.setInt(2,acNo);

                ResultSet rs = ps.executeQuery();

                while(rs.next())
                {
                    String name = rs.getString("Name");
                    int acNo = rs.getInt("Ac_No");
                    float bal = rs.getFloat("Balance");

                    System.out.println("    "+name+"        "+acNo+"        "+bal);
                }

i am getting this exception....plz help me...

error:

java.sql.SQLException: Parameter index out of range (2 > number of parameters, which is 1).

Jens · Accepted Answer · 2017-07-13 07:57:43Z

1

You only have one Parameter to bind. Which has the index one:

So:

ps.setInt(2,acNo);

must be:

ps.setInt(1,acNo);

answered Jul 13, 2017 at 7:57

Jens

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3 Comments

Sohan Kumar Over a year ago

Bt i have provided acNo at 2nd index on the given table... @jens

Sohan Kumar Over a year ago

Is this problem related to the index numbering 9 or parameter ????

access_granted Over a year ago

setInt first parameter relates to the index of "?" expressions in your query. You only have one, so the setInt should bind to "1" only.

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