Question

我有一个案例需要引用另一个数据库用户。我在引用数据库用户名时会硬编码。

SELECT * FROM  eg001t3.DUAL; // example.

有没有办法动态地或基于数据库设置从视图中引用该db用户（eg001t3）？

Answer 1

在pl / sql中，您可以使用EXECUTE IMMEDIATE或DBMS_SQL来动态引用对象。

以EXECUTE IMMEDIATE为例：

SQL> VARIABLE dyn_user VARCHAR2(30);
SQL> EXEC :dyn_user := 'SYS';

PL/SQL procedure successfully completed
dyn_user
---------
SYS
SQL> DECLARE
  2     ln NUMBER;
  3  BEGIN
  4     EXECUTE IMMEDIATE 'SELECT 1
  5                          FROM ' || dbms_assert.schema_name(:dyn_user) 
  6                                 || '.DUAL'
  7        INTO ln;
  8     dbms_output.put_line(ln);
  9  END;
 10  /

1

PL/SQL procedure successfully completed

您还可以使用动态构建的REF CURSOR：

SQL> DECLARE
  2     lc SYS_REFCURSOR;
  3     ln NUMBER;
  4  BEGIN
  5     OPEN lc FOR 'SELECT 1
  6                    FROM ' || dbms_assert.schema_name(:dyn_user) || '.DUAL
  7                   CONNECT BY level <= 2';
  8     LOOP
  9        FETCH lc
 10           INTO ln;
 11        EXIT WHEN lc%NOTFOUND;
 12        dbms_output.put_line(ln);
 13     END LOOP;
 14     CLOSE lc;
 15  END;
 16  /

1
1

如图所示，您可以使用DBMS_ASSERT验证您的输入。

Answer 2

我添加了一个新答案来演示jva建议的其他方法。所有表必须共享一个公共结构（以便Oracle能够在编译时知道视图列的数据类型）。

设定：

-- create 2 schemas
CREATE USER u1 IDENTIFIED BY u1;
CREATE USER u2 IDENTIFIED BY u2;
GRANT RESOURCE TO u1;
GRANT RESOURCE TO u2;

-- one table in each schema
CREATE TABLE u1.t AS 
   SELECT 2 * ROWNUM ID, 'foo' DATA FROM dual CONNECT BY LEVEL <= 5;
CREATE TABLE u2.t AS 
   SELECT 2 * ROWNUM - 1 ID, 'bar' DATA FROM dual CONNECT BY LEVEL <= 5;
GRANT SELECT ON u2.t TO u1;

-- the common structure
CREATE TYPE u1.t_row AS OBJECT (ID NUMBER, DATA VARCHAR2(3));
/
CREATE TYPE u1.t_row_list AS TABLE OF u1.t_row;
/

CREATE OR REPLACE PACKAGE u1.test_pck IS
   schema_name VARCHAR2(30) := 'U1';
   FUNCTION select_t RETURN u1.t_row_list PIPELINED;
END test_pck;
/

--Definition of the pipelined function and the view:
CREATE OR REPLACE PACKAGE BODY u1.test_pck IS

   FUNCTION select_t RETURN u1.t_row_list PIPELINED IS
      l_rc     SYS_REFCURSOR;
      l_id     NUMBER;
      l_data   VARCHAR2(3);
   BEGIN
      OPEN l_rc FOR 'SELECT id, data 
                       FROM ' || dbms_assert.schema_name(schema_name) || '.t';
      LOOP
         FETCH l_rc
            INTO l_id, l_data;
         EXIT WHEN l_rc%NOTFOUND;
         PIPE ROW (u1.t_row(l_id, l_data));
      END LOOP;
      CLOSE l_rc;
   END select_t;

END test_pck;
/

CREATE OR REPLACE VIEW u1.v AS 
SELECT ID, DATA 
  FROM TABLE(u1.test_pck.select_t);

然后，您将在包含模式名称的包中定义全局变量，然后查询视图：

SQL> EXEC u1.test_pck.schema_name := 'U1';

PL/SQL procedure successfully completed
SQL> SELECT * FROM u1.v;

        ID DATA
---------- ----
         2 foo
         4 foo
         6 foo
         8 foo
        10 foo
SQL> EXEC u1.test_pck.schema_name := 'U2';

PL/SQL procedure successfully completed
SQL> SELECT * FROM u1.v;

        ID DATA
---------- ----
         1 bar
         3 bar
         5 bar
         7 bar
         9 bar

Answer 3

可能有更优雅的选项，但您可以动态创建同义词或使用动态SQL / EXECUTE IMMEDIATE。

Answer 4

可能对您有用的另一个选项（取决于您需要执行此操作的应用程序环境）是暂时将您的命名空间更改为感兴趣的架构：

alter session set current_schema = eg001t3;
从whateverTableBelongsToEG001T3中选择*; - 此处不需要架构限定符
alter session set current_schema = ... - 返回到您的连接架构名称

set current_schema不会以任何方式绕过Oracle权限模型 - 您仍然需要在其他模式的感兴趣的表上至少使用SELECT。

如何动态引用另一个数据库用户？

4 个答案: