0

对于一个项目，我们目前进行数据库设计。我们认为我们应该在一张表中使用两个 auto_increment 字段。

表主：

`pid` int(10) NOT NULL auto_increment,
`iid` int(10) NOT NULL auto_increment,
...

要从备用的 auto_incremet 开始，您可以使用ALTER TABLE tbl AUTO_INCREMENT = 100000;这仅适用于整个表 'tbl'。pid 的 auto_increment 应为 50000000，iid 的 auto_increment 应为 80000000

我们希望避免将其拆分为 3 个关系为 master -> table.pid 和 master -> table.iid 的表。

改变桌子不起作用的原因 /* SQL Error (1075): Incorrect table definition; there can be only one auto column and it must be defined as a key */

有可能还是您推荐什么替代方案？

4

2 回答 2

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如果您不能使用两个自动列，我认为您必须重新设计数据库。你到底需要什么？

于 2010-11-05T09:10:52.153 回答
0

我不完全理解您的问题，但您可以使用触发器来维护如下关键值：

drop table if exists grid;
create table grid
(
grid_id int unsigned not null auto_increment primary key,
name varchar(255) not null,
next_token_id int unsigned not null default 0,
next_node_id int unsigned not null default 0
)
engine = innodb;

drop table if exists grid_token;
create table grid_token
(
grid_id int unsigned not null,
token_id int unsigned not null,
name varchar(255) not null,
primary key (grid_id, token_id) -- note clustered PK order (innodb only)
)
engine = innodb;

drop table if exists grid_node;
create table grid_node
(
grid_id int unsigned not null,
node_id int unsigned not null,
name varchar(255) not null,
primary key (grid_id, node_id) -- note clustered PK order (innodb only)
)
engine = innodb;

-- TRIGGERS

delimiter #

create trigger grid_token_before_ins_trig before insert on grid_token
for each row
begin

declare tid int unsigned default 0;

  select next_token_id + 1 into tid from grid where grid_id = new.grid_id;
  set new.token_id = tid;
  update grid set next_token_id = tid where grid_id = new.grid_id;

end#

create trigger grid_node_before_ins_trig before insert on grid_node
for each row
begin

declare nid int unsigned default 0;

  select next_node_id + 1 into nid from grid where grid_id = new.grid_id;
  set new.node_id = nid;
  update grid set next_node_id = nid where grid_id = new.grid_id;

end#

delimiter ;

-- TEST DATA

insert into grid (name) values ('g1'),('g2'),('g3');

insert into grid_token (grid_id, name) values
(1,'g1 t1'),(1,'g1 t2'),(1,'g1 t3'),
(2,'g2 t1'),
(3,'g3 t1');

insert into grid_node (grid_id, name) values
(1,'g1 n1'),(1,'g1 n2'),
(2,'g2 n1'),
(3,'g3 n1'),(3,'g3 n2');

select * from grid;
select * from grid_token;
select * from grid_node;
于 2010-11-05T09:15:35.657 回答