Oracle全局临时表中的保存点
我了解到Oracle全局临时表中的保存点会删除所有数据,但当我在Oracle 11g上测试时,它们的工作方式与堆表类似。有人能解释一下吗Oracle全局临时表中的保存点,oracle,oracle11g,temp-tables,savepoints,Oracle,Oracle11g,Temp Tables,Savepoints,我了解到Oracle全局临时表中的保存点会删除所有数据,但当我在Oracle 11g上测试时,它们的工作方式与堆表类似。有人能解释一下吗 insert into table_1 values('one'); insert into table_1 values('two'); savepoint f1; insert into table_1 values('three'); insert into table_1 values('four'); rollback to f1; -- the
insert into table_1 values('one');
insert into table_1 values('two');
savepoint f1;
insert into table_1 values('three');
insert into table_1 values('four');
rollback to f1;
-- the records in table are 2 records just like heap tables, but I read that
-- savepoints in GTT truncates all the data
你在哪里读到的?我怀疑Oracle SQL引用中没有。因此,解释很简单:该断言的作者没有测试全局临时表的行为。或者您正在阅读其他一些SQL实现的描述,例如DerbyDB 为了完整起见,让我们排除事务或会话范围的角色。以下是两个全局临时表:
create global temporary table gtt1
( col1 varchar2(30) )
ON COMMIT PRESERVE ROWS
/
create global temporary table gtt2
( col1 varchar2(30) )
ON COMMIT DELETE ROWS
/
让我们为具有会话范围的一个运行您的实验:
SQL> insert into gtt1 values('one');
1 row created.
SQL> insert into gtt1 values('two');
1 row created.
SQL> savepoint f1;
Savepoint created.
SQL> insert into gtt1 values('three');
1 row created.
SQL> insert into gtt1 values('four');
1 row created.
SQL> rollback to f1;
Rollback complete.
SQL> select * from gtt1;
COL1
------------------------------
one
two
SQL>
SQL> insert into gtt2 values('five');
1 row created.
SQL> insert into gtt2 values('six');
1 row created.
SQL> savepoint f2;
Savepoint created.
SQL> insert into gtt2 values('seven');
1 row created.
SQL> insert into gtt2 values('eight');
1 row created.
SQL> rollback to f2;
Rollback complete.
SQL> select * from gtt2;
COL1
------------------------------
five
six
SQL>
对于具有事务范围的表,结果相同:
SQL> insert into gtt1 values('one');
1 row created.
SQL> insert into gtt1 values('two');
1 row created.
SQL> savepoint f1;
Savepoint created.
SQL> insert into gtt1 values('three');
1 row created.
SQL> insert into gtt1 values('four');
1 row created.
SQL> rollback to f1;
Rollback complete.
SQL> select * from gtt1;
COL1
------------------------------
one
two
SQL>
SQL> insert into gtt2 values('five');
1 row created.
SQL> insert into gtt2 values('six');
1 row created.
SQL> savepoint f2;
Savepoint created.
SQL> insert into gtt2 values('seven');
1 row created.
SQL> insert into gtt2 values('eight');
1 row created.
SQL> rollback to f2;
Rollback complete.
SQL> select * from gtt2;
COL1
------------------------------
five
six
SQL>
其实这并不奇怪。各国:
“临时表定义的持久化方式与常规表的定义相同”
基本上它们是堆表。区别在于:
- 数据的范围(可见性)
- 用于持久化数据的表空间(全局临时表写入临时表空间)李>
SAVEPOINT f0;
SELECT * FROM v$tempseg_usage; -- Should show nothing for your session
insert into table_1 values('one');
insert into table_1 values('two');
SELECT * FROM v$tempseg_usage; -- Should show a DATA row for your session
savepoint f1;
insert into table_1 values('three');
insert into table_1 values('four');
rollback to f1; -- Undo three and four but preserve one and two
SELECT * FROM v$tempseg_usage; -- Still shows a DATA row for your session
rollback to f0; -- Undo all the inserts
SELECT * FROM v$tempseg_usage; -- row for your session has gone
这很重要的原因是,当您执行正常的删除操作(而不是截断操作)时,表的任何完整扫描仍然必须筛选所有数据块,以查看它们是否包含任何数据。如果在以前某个时间表中有大量数据,那么针对空表的DML可能会导致大量I/O强>
我正试图加快一些代码的运行速度——它将一些内容作为草稿插入一个临时表中,部分原因是它可以连接到一个永久表,并将结果返回给它的调用者。临时表仅用于此例程,因此在例程结束时可以安全地将其清除,但它可能在父事务中被多次调用,因此我无法截断(truncate
是DDL,因此提交事务),但我也无法将其清除,或同一事务中的调用将拾取彼此的行。通过DELETE进行清除会造成相当大的开销,特别是因为表上没有索引,所以针对它进行选择将始终是完全扫描
我正在探索的选项是在例程开始时使用保存点
,执行临时工作,然后在返回结果之前回滚到保存点。另一种选择可能是将例程放在自治事务中,但这意味着将C代码移植到PL/SQL存储过程中,如果临时表需要连接到调用方插入的未提交数据,则无论如何都无法工作
请注意,我是在12c中进行研究的——本版本中对临时表进行了一些改进(请参阅),但我认为这不会影响wrt保存点的行为