为了实现您的目标,您必须使用triggers。没有其他直接的方法可以完成这项任务(我猜)。
我现在确实尝试了一个快速演示:
Create Table SoQuestion (
UserId int,
PostId int,
PostNumber int null
);
CREATE TRIGGER inc_post_num
BEFORE INSERT ON SoQuestion
FOR EACH ROW
set New.PostNumber = (select num
From (select count(*) as num
from SoQuestion
where UserId = New.UserId) as b)
+ 1;
insert into SoQuestion (UserId, PostId) Values (1,1);
insert into SoQuestion (UserId, PostId) Values (1,10);
insert into SoQuestion (UserId, PostId) Values (1,20);
insert into SoQuestion (UserId, PostId) Values (2,1);
insert into SoQuestion (UserId, PostId) Values (2,10);
insert into SoQuestion (UserId, PostId) Values (3,1);
insert into SoQuestion (UserId, PostId) Values (4,1);
select * FROM SoQuestion;
这是我得到的输出:
UserId | PostId | PostNumber |
==============================
1 | 1 | 1 |
1 | 10 | 2 |
1 | 20 | 3 |
2 | 1 | 1 |
2 | 10 | 2 |
3 | 1 | 1 |
4 | 1 | 1 |
这是demo。
在浏览了Auto_Increment 文档后,我找到了另一种不使用触发器来实现此目的的方法。这个想法是关于创建一个Auto_Increment 列并将其与另一列一起添加为PRIMARY KEY。在我们的例子中,它是UserId,AUTO_INCREMENT 是PostNumber,它们都构成了主键。方法是这样的:
Create Table SoQuestion (
UserId int,
PostId int,
PostNumber int NOT NULL AUTO_INCREMENT,
PRIMARY KEY (UserId, PostNumber)
);
insert into SoQuestion (UserId, PostId) Values (1,1);
insert into SoQuestion (UserId, PostId) Values (1,10);
insert into SoQuestion (UserId, PostId) Values (1,20);
insert into SoQuestion (UserId, PostId) Values (2,1);
insert into SoQuestion (UserId, PostId) Values (2,10);
insert into SoQuestion (UserId, PostId) Values (3,1);
insert into SoQuestion (UserId, PostId) Values (4,1);
select * FROM SoQuestion;
这将为我们提供与第一种方式相同的输出:
UserId | PostId | PostNumber |
==============================
1 | 1 | 1 |
1 | 10 | 2 |
1 | 20 | 3 |
2 | 1 | 1 |
2 | 10 | 2 |
3 | 1 | 1 |
4 | 1 | 1 |
这是第二种方式的demo。