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What's mean ORA-25191?

2017-01-24 00:00 344 查看
在参考了Using DBMS_SYS_SQL Package to grant Privilege留言板里有朋友继续问道,出现了以下错误:

ORA-25191: cannot reference overflow table of an index-organized table
ORA-06512: at "SYS.DBMS_SYS_SQL", line 1474
ORA-06512: at line 9
查一下Oracle手册,我们看到:

ORA-25191 cannot reference overflow table of an index-organized table

Cause: An attempt was made to directly access the overflow table of an index-organized table.
Action: Issue the statement against the parent index-organized table containing the specified overflow table.
意思是说,对于Overflow的IOT表,只需要对父表进行授权即可。简单动手自己试试看也可以:

SQL> connect javis/javis
Connected.
SQL> CREATE TABLE TEST_IOT
2        (id NUMBER PRIMARY KEY,
3         C1 VARCHAR2(50),
4         C2 VARCHAR2(10))
5        ORGANIZATION INDEX PCTTHRESHOLD  10 OVERFLOW;

Table created.

SQL> col object_name for a30
SQL> select object_name,object_type from user_objects where object_name like '%IOT%';

OBJECT_NAME                    OBJECT_TYPE
------------------------------ ------------------
SYS_IOT_OVER_7370              TABLE
SYS_IOT_TOP_7370               INDEX
TEST_IOT                       TABLE

SQL> grant select on SYS_IOT_OVER_7370 to scott;
grant select on SYS_IOT_OVER_7370 to scott
*
ERROR at line 1:
ORA-25191: cannot reference overflow table of an index-organized table

SQL>
Ok,那么接下来的就简单了,只要把我们之前的过程简单修改一下:

declare
sqltext varchar2(200);
c integer;
begin
for userlist in (select user_id,username from all_users where username not in ('SYS','SYSTEM','EYGLE')) loop
for tablelist in (select owner,table_name from dba_tables where owner = userlist.username and table_name not like '%IOT_OVER%') loop
sqltext := 'grant all on '||tablelist.owner||'.'||tablelist.table_name ||' to eygle with grant option';
c := sys.dbms_sys_sql.open_cursor();
sys.dbms_sys_sql.parse_as_user( c,sqltext,dbms_sql.native,userlist.user_id);
sys.dbms_sys_sql.close_cursor(c);
end loop;
end loop;
end;
/
这样就可以了.
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