Question

我正在尝试创建一个从表中的最大id开始的序列。

我回答了这个先前提出的问题 How to create sequence using starting value from query?

这是顺序：

CREATE SEQUENCE q9_seq;

SELECT setval('q9_seq', (SELECT MAX(department_id) FROM hr_departments));

即时通讯的结果是：

ORA-00923：在预期错误处找不到FROM关键字。

Answer 1

declare
v_num number;
begin
    select nvl(max(value), 0) + 1
    into v_num
    from table;

    execute immediate 'create sequence drop_me_seq start with ' || v_num;
end;
/

Answer 2

您可以为此使用客户端特定的功能，例如替换变量。在SQL * Plus或SQL Developer中：

column v_start new_value v_start

select nvl(max(department_id), 0) + 1 as v_start from hr_departments;

create sequence q9_seq start with &v_start;

old:create sequence q9_seq start with &v_start
new:create sequence q9_seq start with 43

Sequence Q9_SEQ created.

select q9_seq.nextval from dual;

   NEXTVAL
----------
        43

或者您可以更改现有的顺序以跳过数字，或致电nextval直到达到现有的最大值（as here）。

或者您可以使用一个简单的匿名块：

declare
  l_start number;
begin
  select nvl(max(department_id), 0) + 1 into l_start from hr_departments;

  execute immediate 'create sequence q9_seq start with ' || l_start;
end;
/

PL/SQL procedure successfully completed.

select q9_seq.nextval from dual;

   NEXTVAL
----------
        43

动态SQL语句的生成方式与替换变量方法一样，生成方式为create sequence q9_seq start with 43。

（这两个都假设您的表的当前最高值为42 ...）

setval-在错误的位置找到FROM关键字

2 个答案: