我在外键上有一个列表分区表。因此,如果我想插入一个新实体,则缺少的分区会在插入时引发异常。我以为我是一个很酷的公爵,并使用触发器来创建新分区:-)但是在执行期间分区将不可用。如果你稍等一下,一切正常(但dbms_lock.sleep不会做的伎俩)。
所以这是我的触发器(以及需要的程序) - 注意结束附近的“检查分区部分”
CREATE OR REPLACE PROCEDURE Execute_DDL
(i_sql IN VARCHAR2)
AS
pragma autonomous_transaction;
BEGIN
EXECUTE IMMEDIATE (i_sql);
commit;
END;
/
CREATE OR REPLACE TRIGGER Q_FLD_NEW_PART_TRG
AFTER INSERT
ON Q_FLD
REFERENCING NEW AS NEW OLD AS OLD
FOR EACH ROW
DECLARE
l_cnt number;
l_wait_cnt number := 0;
l_alter varchar2(1000);
l_job_stmt varchar2(1000);
l_job_nr number;
l_job dba_jobs_running%rowtype;
BEGIN
SELECT count(*) INTO l_cnt from user_tables
where table_name = 'QUOTE' and partitioned = 'YES';
if l_cnt <= 0 then return; end if;
l_alter := 'ALTER TABLE QUOTE ADD PARTITION QUOTE_PART_' || :new.name || ' VALUES (' || :new.id || ')';
l_job_stmt := 'begin Execute_DDL (''' || l_alter || '''); end;';
DBMS_JOB.SUBMIT (job => l_job_nr, what => l_job_stmt);
if l_job_nr is null then
raise_application_error(-20000, 'Partition Job Creation failed!', true);
end if;
-- wait for job to complete
while l_job_nr is not null loop
l_wait_cnt := l_wait_cnt +1;
if l_wait_cnt > 30 then raise_application_error(-20000, 'pratition creation timed out!'); end if;
begin
select * into l_job from dba_jobs_running where job = l_job_nr;
if l_job.failures >0 then
raise_application_error(-20000, l_job_stmt, true);
end if;
sys.dbms_lock.sleep(2);
exception when no_data_found then
l_job_nr := null; -- job completed
end;
end loop;
-- check partition available
/* this will lead into a "no data found" exception.
so i can not use the new partition immediatly. why??
sys.dbms_lock.sleep(2);
select count(*) into l_cnt
from user_objects
where object_type = 'TABLE PARTITION'
and subobject_name = 'QUOTE_PART_' || upper(:new.name);
if l_cnt <= 0 then
raise_application_error(-20000, 'Partition creation falied/timed out: ' || 'QUOTE_PART_' || :new.name, true);
end if;
*/
exception when others then
raise_application_error(-20000, l_job_stmt, true);
END q_fld_new_part_trg;
/
有人想要解决这个问题吗?我在Linux上使用11gR2 64位
答案 0 :(得分:6)
由于您使用的是Oracle 11.2,是否有理由不在此处使用interval partitioning?假设ID
是一个数字列,这样的事情会告诉Oracle每次插入新的ID
值时都会创建一个新的分区。
SQL> create table interval_table(
2 id number,
3 value varchar2(10)
4 )
5 partition by range(id)
6 interval( 1 )
7 (
8 partition initial_partition values less than (2)
9 );
Table created.
SQL> insert into interval_table( id, value )
2 values( 1, 'Initial' );
1 row created.
SQL> insert into interval_table( id, value )
2 values( 10, 'New' );
1 row created.