我希望创建一个简单的序列:
CREATE SEQUENCE supplier_seq
MINVALUE 1
START WITH 3
INCREMENT BY 4
CACHE 20000;
但是,我想使用由Java端实现的某些因素确定的函数对增量值进行半随机化。
我认为测试的第一步可能看起来像这样:
CREATE SEQUENCE supplier_seq
MINVALUE 1
START WITH 3
INCREMENT BY myfunction
CACHE 20000;
我尝试过:
CREATE SEQUENCE supplier_seq
MINVALUE 1
START WITH 3
INCREMENT BY (declare
rtrnt number;
begin
rtrnt :=semiRandomize();
end;
)
CACHE 20000;
我意识到这很荒谬..但是必须有某种方法可以做这样的事情。有指针吗?
答案 0 :(得分:0)
这个怎么样?并不是真正的序列,但是-可能适合您的需求。它是一种选择随机数并将其存储在具有主键的表中的功能。如果该号码已被使用,则会被跳过。该功能检查是否已使用所有数字;如果是这样,则会引发错误。
在此示例中,我创建了一个5位数的“序列”(这样它很快就会失败)。
表和功能:
SQL> create table t_seq (supseq number constraint pk_tseq primary key);
Table created.
SQL> create or replace function f_supseq
2 return number
3 as
4 l_range number := 5; -- number of values in a "sequence"
5 l_seq number; -- a new "sequence" number
6 l_cnt number; -- number of used numbers
7 pragma autonomous_transaction;
8 begin
9 select count(*) into l_cnt from t_seq;
10 if l_cnt < l_range then
11 -- there are still some available numbers so - let's get them
12
13 -- don't let anyone mess with the table
14 lock table t_seq in exclusive mode;
15 while l_seq is null loop
16 begin
17 insert into t_seq (supseq) values
18 (round(dbms_random.value(1, l_range)))
19 returning supseq into l_seq;
20 exception
21 when dup_val_on_index then
22 -- that number has already been used; skip it
23 null;
24 end;
25 end loop;
26 commit;
27 else
28 raise_application_error(-20001, 'No more available numbers');
29 end if;
30
31 return l_seq;
32 end;
33 /
Function created.
获取随机的“序列”值:
SQL> select f_supseq from dual;
F_SUPSEQ
----------
2
SQL> select f_supseq from dual;
F_SUPSEQ
----------
4
SQL> select f_supseq from dual;
F_SUPSEQ
----------
1
SQL> select f_supseq from dual;
F_SUPSEQ
----------
3
SQL> select f_supseq from dual;
F_SUPSEQ
----------
5
SQL> select f_supseq from dual;
select f_supseq from dual
*
ERROR at line 1:
ORA-20001: No more available numbers
ORA-06512: at "SCOTT.F_SUPSEQ", line 28
表内容:
SQL> select * From t_seq;
SUPSEQ
----------
1
2
3
4
5
SQL>