我当前的序列号是 203 我需要跳到 1203 而不丢弃序列。
3 回答
3
您可以在所需的 DB 模式上创建这样的过程,如下所示:
SQL>Create or Replace Procedure Pr_Set_Sequence( i_seq_name varchar2, i_val pls_integer ) is
v_val pls_integer;
begin
for c in (
Select u.sequence_name seq
From User_Sequences u
Where u.sequence_name = upper(i_seq_name)
)
loop
execute immediate 'select '||i_seq_name||'.nextval from dual' INTO v_val;
execute immediate 'alter sequence '||i_seq_name||' increment by ' ||
to_char(-v_val+i_val) || ' minvalue 0';
execute immediate 'select '||i_seq_name||'.nextval from dual' INTO v_val;
execute immediate 'alter sequence '||i_seq_name||' increment by 1 minvalue 0';
end loop;
end;
并使用所需的值调用(在我的情况下,创建一个名为的新值my_seq):
SQL> create sequence my_seq;
Sequence created
SQL> select my_seq.nextval from dual;
NEXTVAL
----------
1
SQL> begin
2 pr_set_sequence('my_seq',1203);
3 end;
4 /
PL/SQL procedure successfully completed
SQL> select my_seq.currval from dual;
NEXTVAL
----------
1203
于 2018-07-20T14:10:32.583 回答
3
首先改变序列增量,给定你说的203到1203,我加1000,酌情调整。
ALTER SEQUENCE yourSequence INCREMENT BY 1000;
然后请求一个值
SELECT yourSequence.NextVal FROM dual;
然后将其改回以 1 递增(假设它首先是 1)
ALTER SEQUENCE yourSequence INCREMENT BY 1;
如果序列正在使用中,您真的不想这样做 - 因为它可能会跳跃数千。
于 2018-07-20T13:45:10.270 回答
0
只需执行一个 WHILE 循环,从序列中进行选择,直到达到您需要的值。它相当快。
declare
v_sequence_value number;
begin
while v_sequence_value <= 1203 loop
select my_sequence.nextval into v_sequence_value from dual;
end loop;
end;
/
于 2018-07-23T20:44:41.570 回答