如何从不同的表中获取值并将其传递给触发器？

Question

我有3张桌子：

event:
---------------------
event_id | uid | date
---------------------

event_invitees
----------------------
invitees_id | event_id
---------------------- 

notification
-----------------
uid | invitees_id
-----------------

我创建了一个触发器，在'event_invitees'上插入后将插入'通知'表中

delimiter |
create trigger eventNotificationTrigger after insert on event_invitees
for each row 
begin
declare id int(12);
set id = (select uid from event where event_id=new.event_id);
insert into notification (invitees_id,uid) select invitees_id,@id from event_invitees where event_id=new.event_id;
end;
|

但它无法从事件表中获取uid。那么如何从不同的表中获取不同的值并将它们插入触发器中的一个表？

Answer 1

删除@

insert into notification (invitees_id,uid) 
select invitees_id, @id from event_invitees where ...
                    ^-------------------------------------here

使用declare创建的变量不会被@引用。只有您使用set创建的那个。例如：

declare id int(12);
select 1 into id;
select id;

set @id = (select 1);
select @id;