我希望从Java中的SQL * Plus以SYS
身份连接到Oracle。但我无法连接。
但我能够以名为SCOTT
的用户身份进行连接。我的代码段如下:
public static void test_script () {
String fileName = "@t.sql";
//t.sql contains "show user" command
String sqlPath = "D:\\";
String sqlCmd = "sqlplus";
// String arg1 = "scott/tiger@(DESCRIPTION=(ADDRESS=(PROTOCOL=TCP)(Host=hostname)(Port=PORT ID))(CONNECT_DATA=(SID=SID)))";
String arg1 = "sys as sysdba/tiger@(DESCRIPTION=(ADDRESS=(PROTOCOL=TCP)(Host=hostname)(Port=PORT ID))(CONNECT_DATA=(SID=SID)))";
//String arg1="/ as sysdba";
String arg2= fileName;
//String arg2="conn /as sysdba";
try {
String line;
ProcessBuilder pb = new ProcessBuilder(sqlCmd, arg1,arg2);
Map<String, String> env = pb.environment();
env.put("VAR1", arg1);
env.put("VAR2", arg2);
//env.put("VAR3", arg3);
pb.directory(new File(sqlPath));
pb.redirectErrorStream(true);
Process p = pb.start();
BufferedReader bri = new BufferedReader
(new InputStreamReader(p.getInputStream()));
BufferedReader bre = new BufferedReader
(new InputStreamReader(p.getErrorStream()));
while ((line = bri.readLine()) != null) {
System.out.println(line);
}
bri.close();
while ((line = bre.readLine()) != null) {
System.out.println(line);
}
bre.close();
System.out.println("\n\n\n");
System.out.println("Done.");
}
catch (Exception err) {
err.printStackTrace();
}
}
}
当我尝试运行此代码时,我发现此错误:
SQL*Plus: Release 11.2.0.1.0 Production on Thu Apr 10 11:08:59 2014
Copyright (c) 1982, 2010, Oracle. All rights reserved.
SQL*Plus: Release 11.2.0.1.0 Production
Copyright (c) 1982, 2010, Oracle. All rights reserved.
Use SQL*Plus to execute SQL, PL/SQL and SQL*Plus statements.
Usage 1: sqlplus -H | -V
-H Displays the SQL*Plus version and the
usage help.
-V Displays the SQL*Plus version.
Usage 2: sqlplus [ [<option>] [{logon | /nolog}] [<start>] ]
...
...以及其他SQL * Plus'使用'信息。
我是否提供了错误的arg1
参数,或者是否有任何其他方式通过Java在Oracle中连接为SYS
。
答案 0 :(得分:3)
您将所有连接信息作为单个值传递;从命令行等效于此:
sqlplus "sys as sysdba/tiger@<connect_string>"
将获得与打印SQL * Plus登录帮助相同的响应。您的密码也在错误的位置,但它没有那么远。从命令行开始,这将起作用:
sqlplus "sys/tiger" "as" "sysdba@<connect_string>"
所以你需要将5个参数传递给ProcessBuilder
,例如:
String sqlCmd = "sqlplus";
String arg1 = "sys/tiger";
String arg2 = "as";
String arg3 = "sysdba@(DESCRIPTION=(ADDRESS=(PROTOCOL=TCP)(Host=hostname)(Port=PORT ID))(CONNECT_DATA=(SID=SID)))";
String arg4 = fileName;
...
ProcessBuilder pb = new ProcessBuilder(sqlCmd, arg1, arg2, arg3, arg4);
仅当您的环境配置为允许远程连接为sysdba
时,这仍然有效。做任何事情sys
应该是非常罕见的,并且让你想要以sys
运行的脚本看起来不同寻常,以至于Java包装器看起来有点矫枉过正 - 并使你看起来可能像{{{常规地,这不是一个好主意 - 但也许这只是一个学习练习。
答案 1 :(得分:1)
我通过点击找到答案并尝试连接字符串。
如果想要以sysdba / sysoper身份连接,则连接字符串应为:
public static void test_script () {
String fileName = "@t.sql";
String sqlPath = "D:\\";
String sqlCmd = "sqlplus";
// IP_address,portid and sid are variables to be entered and t.sql is the file to be read .It contains show user command
String arg3 = "sys/oracle123@(DESCRIPTION=(ADDRESS=(PROTOCOL=TCP)(Host=IP_address)(Port=portid))(CONNECT_DATA=(SID=sid))) as sysdba";
String arg4= fileName;
try {
String line;
ProcessBuilder pb = new ProcessBuilder(sqlCmd,arg3,arg4);
Map<String, String> env = pb.environment();
env.put("VAR3", arg3);
env.put("VAR4", arg4);
pb.directory(new File(sqlPath));
pb.redirectErrorStream(true);
Process p = pb.start();
BufferedReader bri = new BufferedReader(new InputStreamReader(p.getInputStream()));
BufferedReader bre = new BufferedReader(new InputStreamReader(p.getErrorStream()));
while ((line = bri.readLine()) != null) {
System.out.println(line);
}
bri.close();
while ((line = bre.readLine()) != null) {
System.out.println(line);
}
bre.close();
System.out.println("\n\n\n");
System.out.println("Done.");
} catch (Exception err) {
err.printStackTrace();
}
}
答案 2 :(得分:0)
认为您的arg1应如下所示:
String arg1 = "scott as sysdba/<syspwd>@(DESCRIPTION=(ADDRESS=(PROTOCOL=TCP)(Host=hostname)(Port=PORT ID))(CONNECT_DATA=(SID=SID)))";
答案 3 :(得分:0)
您可以像下面那样从Java轻松调用sqlplus。
String stringCommand =
"sqlplus " + dbUser + "/" + dbPassword + "@(description=(address=(protocol=TCP)" +
"(host=" + dbHost + ")(port=" + dbPort + "))(connect_data=(service_name=" + dbName + "))) " +
"@" + sqlScriptFile + "";
int exitCode;
Runtime rt = Runtime.getRuntime();
Process process = null;
try {
process = rt.exec(stringCommand);
exitCode = process.waitFor();
} finally {
process.destroy();
}