Reputation: 3
I have a music database so I'll need a search function, now when entering into the search bar, I get the error : "Invalid argument in JDBC call: parameter index out of range: 1"
public static Song getSongByName(String name)
{
String sql =
"SELECT songID FROM Song "+
"WHERE name = '?'; ";
Connection conn = Connections.getConnection();
Song erg = null;
try
{
PreparedStatement pstmt = conn.prepareStatement(sql);
pstmt.setString(1,name);
ResultSet rs = pstmt.executeQuery();
rs.next();
int id = rs.getInt (1);
name = rs.getString (2);
erg = new SongImpl(id,name);
rs.close();
pstmt.close();
}
catch(SQLException exc)
{
System.err.println("Fehler: in SQL-Aufruf");
System.err.println("["+sql+"]");
exc.printStackTrace();
System.exit(6);
}
Connections.putConnection(conn);
return erg;
}
Upvotes: 0
Views: 115
Reputation: 475
It seems like you are selecting just one column instead of two.
The correct SQL query would be:
SELECT songID, songName FROM Song WHERE name = '?';
Try this:
public static Song getSongByName(String name)
{
String sql =
"SELECT songID, songName FROM Song "+
"WHERE name = ?";
Connection conn = Connections.getConnection();
Song erg = null;
try
{
PreparedStatement pstmt = conn.prepareStatement(sql);
pstmt.setString(1,name);
ResultSet rs = pstmt.executeQuery();
rs.next();
int id = rs.getInt(1);
String name = rs.getString(2);
erg = new SongImpl(id, name);
rs.close();
pstmt.close();
}
catch(SQLException exc)
{
System.err.println("Fehler: in SQL-Aufruf");
System.err.println("["+sql+"]");
exc.printStackTrace();
System.exit(6);
}
Connections.putConnection(conn);
return erg;
}
Let me know if that helped :)
Upvotes: 1
Reputation: 65
Look onto examples https://docs.oracle.com/javase/tutorial/jdbc/basics/prepared.html
Try to remove single quotes:
String sql =
"SELECT songID FROM Song "+
"WHERE name = ?; ";
You should use string parameters without quotes because your statement interpreted as a SQL query without parameters.
Upvotes: 0