Java Servlet & MySQL

Question

What's wrong with this code? It gives me "brand" = null.

public String getBrand(@WebParam(name = "regNr") String regNr) throws SQLException { String brand=null; Connection con; Statement st; ResultSet rs; String url = "jdbc:mysql://db4free.net:3606/cars"; String user = "cars"; String password = ""; try { con = (Connection) DriverManager.getConnection(url, user, password); st = (Statement) con.createStatement(); rs = st.executeQuery("SELECT * FROM Cars WHERE nr= '" + regNr + "'"); if (rs.next()) { brand=rs.getString("brand"); } } catch (SQLException ex) { } return brand; }

I want to display the value of brand from database. Is it a problem with connection?

Asaph · Accepted Answer · 2013-11-30 22:43:44Z

Either the query isn't returning any rows or a SQLException is being thrown. Add something to your catch block to discover what the exception is.

BTW: You should be calling close() on your ResultSet, Statement and Connection instances. I suggest putting those calls within a finally block. Additionally, your code is vulnerable to an attack called SQL injection. Use a PreparedStatement with parameters to repair this problem.

kgautron · Accepted Answer · 2013-11-30 22:46:52Z

There probably is an exception in the code but you don't see it because of the catch with nothing inside. Add a log of the exception inside it, to reveal the exception and see the problem.

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