Question

I am trying to write java code to access a table ‘customer’ with columns ‘customer_id’, ’email’, ‘deliverable’, and ‘create_date’

I have

        Connection conn = DriverManager.getConnection(connectionUrl, connectionUser, connectionPassword);
        Statement constat = conn.createStatement();
        String query = "SELECT * FROM customer WHERE customer_id LIKE " + customerId;
        ResultSet rtn = constat.executeQuery(query);
        Customer cust = new Customer(rtn.getInt("customer_id"), rtn.getString("email"),  rtn.getInt("deliverable"), rtn.getString("create_date"));
        conn.close();
        return cust;

I am receiving the error:

java.sql.SQLException: Before start of result set

As far as I can tell, my error is in the line where I am creating a new Customer object, but I cannot figure out what I am doing wrong. Can anyone offer me some help? Thanks!

Answer 1

You must always go to the next row by calling resultSet.next() (and checking it returns true), before accessing the data of the row:

Customer cust = null;
if (rtn.next()) {
    cust = new Customer(rtn.getInt("customer_id"), 
                        rtn.getString("email"), 
                        rtn.getInt("deliverable"), 
                        rtn.getString("create_date"));
}

Note that you should also

• use prepared statements instead of String concatenation to avoid SQL injection attacks, and have more robust code
• close the connections, statements and resultsets in a finally block, or use the try-with-resources construct if using Java 7

Read the JDBC tutorial