32

我收到此错误：

java.sql.SQLException: ORA-28009: connection as SYS should be as SYSDBA or SYSOPER

怎么修？（我需要SYS）。谢谢。

4

9 回答 9

41

试试这个 ：

import java.sql as jsql
import java.lang as lang
driver, url, user, passwd = (
"oracle.jdbc.driver.OracleDriver",
"jdbc:oracle:thin:@localhost:1234:xxx1",
"sys as sysdba",
"xxx1")
 lang.Class.forName(driver)
 c = jsql.DriverManager.getConnection(url,user,passwd)
于 2012-04-11T18:04:20.893 回答
14

已经有了答案，

您正在尝试连接，sys但服务器允许

任何一个

sys as sysdba

或者

sys as sysoper

只需将用户参数更改为上面的任何一个

user='sys as sysdba'

或者

user='sys as sysoper'
于 2013-12-05T05:50:38.597 回答
8

此代码有效

String driverName = "oracle.jdbc.driver.OracleDriver";
Class.forName(driverName).newInstance();
String nameForConnect = "sys as sysdba";
String pass = "password";
String url = "jdbc:oracle:thin:@192.168.0.1:1521:ORCL";
Connection conn = DriverManager.getConnection(url, nameForConnect, pass);
于 2012-04-11T17:56:34.397 回答
5

如果您尝试像这样连接到数据库：connect SYS/<password>您使用的语法不再有效（Oracle 9i 之后）。

而是尝试如下连接：

connect SYS/<password> as SYSDBA or connect SYS/<password> as SYSOPER
于 2012-04-11T07:29:43.360 回答
1

您可以使用 OracleDataSource 对象吗？

public class Database {    
    static OracleDataSource ods;    
    public static Connection openConnection(String URL, String user, String password,     String option) throws SQLException
    {
            Connection conn = null;
            Properties properties = new Properties();
            properties.put("user", user);
            properties.put("password", password);

            ods = new OracleDataSource();
            ods.setURL(URL);

            if(option != null)
            {
                properties.put("internal_logon", option);
            }

            ods.setConnectionProperties(properties);
            conn = ods.getConnection();

            return conn;
    }
}

并这样称呼它：

Connection con = null;    
con = Database.openConnection("YourJDBCConnectionURL", "YourSYSUser", "YourSYSPassword", "sysdba");
于 2012-04-12T11:16:52.200 回答
1

如果要将数据库与“sys”以外的用户连接为“sysdba”，则必须将驱动程序从“thin”更改为“oci”才能成功连接。

try {
        Class.forName("oracle.jdbc.driver.OracleDriver");
        String DB_URL="jdbc:oracle:oci:@localhost:1521:orcl";
        OracleDataSource ds1=new OracleDataSource();
        Properties prop1 = new Properties();
        prop1.setProperty("user","ravi");
        prop1.setProperty("password","******");
        prop1.setProperty("internal_logon","sysdba");
        ds1.setConnectionProperties(prop1);
        ds1.setURL(DB_URL);
        OracleConnection conn1 = (OracleConnection)ds1.getConnection();
        Statement stmt = conn1.createStatement();
        ResultSet rs = stmt.executeQuery("select * from dba_users");
        while (rs.next())
            System.out.println(rs.getString(1));
        conn1.close();
    } catch (Exception e) {
        System.out.println(e);
    }
于 2017-06-10T10:54:56.040 回答
1 
        /*It works for me*/
        /*also oci and thin is important if you want to connect with database which is installed in your clien (Computer) add oci if you want to install to server add thin*/

        String dbURL2 = "jdbc:oracle:oci:@172.16.24.123:1521:XE";
        String username = "sys as sysdba";
        String password = "XX Change it to your system password";
        
        try {
        Connection connection = DriverManager.getConnection(dbURL2, username, password);
            System.out.println("Connected to Oracle data");
        
    } catch (SQLException e) {
            System.out.println("Opps ! error");
            e.printStackTrace();
        
    }
于 2020-11-23T22:29:12.170 回答
0

您需要将 sysdba 与用户字符串参数一起放入，例如

String user="sys as sysdba"
于 2018-01-11T12:36:28.553 回答
0

当我遇到同样的异常时，我的两分钱。以下为我做了诀窍：

           try (Connection conn = DriverManager.getConnection(
                "jdbc:oracle:thin:@myserverip:1521:XE", "sys as sysdba", "Hello123")) {   
           // this worked too - "jdbc:oracle:thin:@myserverip:1521:XE", "system"
              if (conn != null) {
                  System.out.println("Connected to the database!");
              } else {
                System.out.println("Failed to make connection!");
              }
 
            } catch (SQLException e) {
              System.err.format("SQL State: %s\n%s", e.getSQLState(), e.getMessage());
            } catch (Exception e) {
            e.printStackTrace();
           }
于 2020-12-08T21:38:20.757 回答