java - 如何在Java中注册Raw类型的OutParameter

Question

我有一个名为 Enter 的 Oracle 过程，它返回 RAW 类型作为结果。当我想注册名为“ret”的参数时，出现此错误:

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

这是我的java方法和导入语句:

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;

}

Java中如何获取RAW类型的输出参数？预先感谢!

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

您有参数名称，而不是参数类型。

尝试仅使用:

cstmt.registerOutParameter(2,OracleTypes.RAW)