use*_*330 9 oracle privileges schema grant synonym
我需要帮助了解用户在指向另一个(不同的)架构对象时创建SYNONYM所需的授权/特权.
当我尝试以下操作时,我得到ora-01031权限不足,显然我失踪并且未能应用其他所需权限.我尽可能地搜索,但找不到任何特定于跨模式同义词的东西.
CREATE USER test IDENTIFIED BY pw DEFAULT TABLESPACE USERS TEMPORARY TABLESPACE TEMP;
ALTER USER test IDENTIFIED BY pw;
GRANT CONNECT, RESOURCE TO test;
-- ... create a bunch of stuff in test...
CREATE USER READWRITE IDENTIFIED BY pw DEFAULT TABLESPACE USERS TEMPORARY TABLESPACE TEMP;
ALTER USER READWRITE IDENTIFIED BY pw;
GRANT CONNECT, RESOURCE TO READWRITE;
GRANT SELECT ON GDACS.FIXALARMS TO PUBLIC;
GRANT UPDATE, INSERT ON GDACS.FIXALARMS TO READWRITE;
CONNECT READWRITE/pw;
CREATE SYNONYM FIXALARMS for test.FIXALARMS;
ORA-01031 insufficient privileges
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Ale*_*ole 14
用于该文档的CREATE SYNONYM
命令包括:
先决条件
要在自己的架构中创建专用同义词,您必须具有
CREATE SYNONYM
系统特权.要在另一个用户的架构中创建私有同义词,您必须具有
CREATE ANY SYNONYM
系统特权.要创建
PUBLIC
同义词,您必须具有CREATE PUBLIC SYNONYM
系统特权.
您正在尝试在READWRITE
自己的架构中创建一个私有同义词,因此您必须这样做:
GRANT CREATE SYNONYM TO READWRITE;
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同义词指向的对象位于不同的模式中,但这与此无关.