How to registerOutParameter of type Raw in Java

Question

I have Oracle procedure named enter that returns RAW type as result.When I want to register out parameter named 'ret' I get this error :

 java.sql.SQLException: Parameter Type Conflict: sqlType=-2

This is my java method and import statement:

import org.apache.openjpa.jdbc.sql.Raw;

public Raw enter(int pid) {
    Connection connection;
    CallableStatement cstmt;
    try {
        connection = cscadaDataSource.getConnection();
        connection.setAutoCommit(false);
        cstmt = connection.prepareCall("{ CALL process_logging.PROCESSMONITOR_ENTER(?,?) }");
        cstmt.setInt(1, pid);
        cstmt.registerOutParameter(2,OracleTypes.RAW , "ret");
        cstmt.execute();

        Raw result = (Raw) cstmt.getObject(2);

        connection.commit();
        return result;
    } catch (SQLException e) {
        // TODO Auto-generated catch block
        e.printStackTrace();
    }
    return null;

}

How can I get this output parameter of RAW type in Java? Thanks in advance!

Answer 1

void registerOutParameter(int parameterIndex, int sqlType, String typeName) throws SQLException:

Parameters:
parameterIndex - the first parameter is 1, the second is 2,...
sqlType - a value from Types
typeName - the fully-qualified name of an SQL structured type

You have parameter name, not parameter type.

Try use just:

cstmt.registerOutParameter(2,OracleTypes.RAW)